PRACTICALS OF MEDICAL CHEMISTRY - Kauno ...julivan/medical-chemistry... · Web view10 April 18 10. Carbonyl compounds – aldehydes, ketones and acids. Their properties – specific

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Department of BiochemistryLithuanian University of Health Sciences

PRACTICALS OF MEDICAL CHEMISTRYfor Students of Medicine and Odontology

Kaunas, 2013

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SCHEDULE OF MEDICAL CHEMISTRY LECTURES FOR MEDICINE AND ODONTOLOGY 1st YEAR STUDENTS

( Spring semester, 2013)

The lectures take place at 800–930

N Date Title of lectures Location1 March 28 1. Water and solutions. Prof Ramunė Morkūnienė Room 4272 March 29 2. Properties of buffers. Prof Ramunė Morkūnienė Room 427

March 30 - April 7

Easter holiday

3 April 8 3. Basics of colloidal chemistry. Prof Laima Ivanovienė Room 427

4 April 9 4. Heterogenic processes. Prof Artūras Kašauskas Room 427

5 April 10 5. Coordination compounds. Prof Artūras Kašauskas The Small auditorium

6 April 11 6. Thermodynamics. Prof Artūras Kašauskas Room 427

7 April 12 7. Chemical kinetics and catalysis. (for Medical students only) Prof Laima Ivanovienė

Room 427

April 13April 14

8 April 15

April 16 CONTROL TEST I

9 April 17 8. Classes of organic compounds. Prof Ramunė Morkūnienė

The Small auditorium

10 April 18 9. Isomers. (for Medical students only) Prof Artūras Kašauskas Room 427

11 April 19 10. Alcohols, aldehydes, carboxylic acids. Prof Artūras Kašauskas

Room 427

April 20April 21

12 April 22 11. Carbohydrates. Prof Ramunė Morkūnienė Room 427

13 April 23 12. Nucleotides and nucleic acids. Prof Artūras Kašauskas Room 427

14 April 24 13. Amino acids and proteins. Prof Laima Ivanovienė

The Big auditorium

15 April 25 14. Fatty acids and lipids. Prof Laima Ivanovienė Room 427April 26 CONTROL TEST IIApril 27April 28April 29April 30 FINAL EXAM

Head of the Biochemistry department Prof. L. Ivanovienė

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SCHEDULE OF MEDICAL CHEMISTRY LABORATORY WORK FOR 1st YEAR MEDICINE AND ODONTOLOGY STUDENTS

(Spring semester, 2013)Laboratory practicals take place in laboratory124 945–1145 for groups 31, 32, 36 MF R.Morkūnienė/S.Dubickienė1200–1400 for groups 29, 33 MF L.Ivanovienė1400-1600 for groups 30, 34, 35 MF A.Kašauskas/S.Dubickienė1600-1800 for groups 13-15 OF students A.Kašauskas/S.DubickienėN Date Title of laboratory practicals1 March 28 1. Principles of safe work in a chemical laboratory.

Preparation of solutions. Percent concentration. 2 March 29 2. Preparation of molar solutions: molar concentration.

March 30 - April 7

Easter holiday

3 April 8 3. Volumetric analysis. Acid-base titration.

4 April 9 4. Preparation of buffer solutions.Buffering capacity and its determination.

5 April 10 5. Preparation of colloidal solutions. Effect of ions on coagulation of colloidal solutions.

6 April 11 6. High-molecular mass compounds and their properties: jellynation.

7 April 12 7. Coordination-compound based determination of water hardness.

April 13April 14

8 April 15 8. Last minute arrangements.

April 16 CONTROL TEST I.

9 April 17 9. Chemical catalysis and thermodynamics of chemical processes. Examples of reaction rate changes, effect of a catalyst, equilibrium shift.Determination of Michaelis-Menten constant of an enzyme from experimental data (Dry practical). (This day laboratory works are for Medical students only)

10 April 18 10. Carbonyl compounds – aldehydes, ketones and acids. Their properties – specific reactions of functional groups. Esters and lipids. Determination of products of lipid peroxidation.

11 April 19 11. Basics of organic compounds. Some chemical properties of alcohols, phenols and amines.

April 20

April 2112 April 22 12. Chemical properties of mono- and polysaccharides.13 April 23 13. Nucleotides and nucleic acids: characteristic reactions. Specific reactions for products

of nucleic acid hydrolysis.14 April 24 14. Specific chemical reactions for amines, amino acids and proteins. Estimation of

molecular mass of a protein from the gel chromatography data (Dry practical). Estimation of molecular mass of a protein by the electrophoresis method (Dry practical).

15 April 25 15. Last minute arrangements.April 26 CONTROL TEST IIApril 27April 28April 29 Repetition of control tests if failedApril 30 FINAL EXAM

Head of the Biochemistry department Prof. L. Ivanovienė

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Questions for the control test N1 - 2013

1. Characteristics of water molecule. Distribution of charges. 2. Hydrophylic, hydrophobic and amphipatic compounds. 3. Types of real solutions. Electrolytes and non-electrolytes. 4. Calculation of percent concentration. Calculation of molarity. 5. Osmolarity of real solutions. Isotonic, hypertonic and hypotonic solutions. 6. Electrolytic dissociation of weak electrolytes. Principle of Le Chatelier. 7. Dissociation of water - pH. Calculation of pH as the function of -log[H+]. Calculation of pOH 8. pK - meaning and a way of its determination. 9. Buffering capacity of buffers. What compounds can act as buffers in solutions.10. Bicarbonate buffer system of human blood. Know how to calculate its efficiency using

Henderson-Hasselbalch equation. 11. Properties of colloidal solutions. Composition of micelle. 12. Coagulation of colloidal solutions. What factor is responsible for resistance of colloidal solution

against coagulation. 13. Properties of biopolymer solutions. 14. Solubility and solubility product. Heterogeneous equilibrium. What factors promote shifting of the

heterogeneous equilibrium?15. Adsorption and absorption. Factors which influence adsorption process.

Ion-exchange adsorption.16. Coordination compounds, their properties.17. Complexones. Characteristic of complexones and some examples of them (EDTA, Trilon B). 18. Systems in the thermodynamics.19. First law of thermodynamics.20. Second law of thermodynamic.

Questions to the 2nd control test on Medical Chemistry - 2013

1. Functional groups.2. Sigma and pi bonds.3. Important reactions of organic compounds4. Level of organic compound oxidation5. Naming of organic compounds by IUPAC6. Chemical properties of alcohols and phenols.7. Reactions for identification of aldehyde group.8. Saponification.9. Classification of monosaccharides.10. Structure of main monosaccharides and disaccharides. Anomeric C atom.11. Polysaccharides: structure and types of bonds.12. Physical and chemical properties of saturated fatty acids. Structural formulas of palmitic and stearic

acids.13. Physical and chemical properties of unsaturated fatty acids. Omega-classification. Structural formulas

of linolic and linolenic acids.14. Fats and oils. Structure of glycerophospholipids.15. Properties of cholesterol16. Formation of peptide bond.17. Acid-base properties of amino acids.18. Nitrogen bases, nucleosides and nucleotides: their nomenclature and chemical structure.19. Chemical structure of nucleic acids and its relation to the appropriate biological function.20. Hydrogen bonds between molecules of organic compounds. Solubility in water.

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QUESTIONS FOR FINAL EXAMINATION ON MEDICAL CHEMISTRY2012-2013 academic year

1. Characteristics of water molecule. Hydrogen bond. 2. Hydrophylic, hydrophobic and amphipatic compounds.3. Electrolytes and non-electrolytes.4. Percent concentration (m/m %, v/v %, m/v %). Molar concentration.5. Colligative properties. Osmotic pressure, osmolarity.6. Electrolytic dissociation of weak electrolytes. Le Chatelier's principle.7. Index of acidity (pH).8. Buffers. Buffering capacity. pH calculations with buffers: the Henderson-Hasselbalch equation.9. Carbonic acid-bicarbonate buffer system and phosphate buffer system: principles of their action.10. Real solutions, colloidal solutions and low dispersion systems.11. High-molecular mass compounds, properties of high molecular mass compound solutions.12. Coagulation of colloidal solutions: its kinetics and factors, affecting coagulation.13. Heterogeneous processes. Heterogeneous equilibrium. Examples of the heterogeneous processes occurring in the

human body.14. Adsorption and absorption. Adsorption when adsorbent is liquid substance: surface-active substances, aeroembolism

(decompression sickness).15. Coordination compounds, the structure of coordination compounds. 16. Dissociation of coordination compounds, instability constant of coordination compounds.17. Chelates, their structure. Complexones.18. Hess’s law.19. Entropy and enthalpy.20. Gibbs free energy.21. Exergonic and endergonic reactions.22. Reaction rate. Rate law.23. Effect of temperature on reaction rate.24. Effect of catalysts on chemical reaction: on activation energy, reaction rate and equilibrium constant.25. Sigma bonds in organic compounds26. Pi bonds in organic compounds27. Functional groups.28. Level of organic compounds oxidation.29. Structural isomers.30. Cis- and trans-isomers.31. Chiral carbon atom.32. L and D-isomers.33. Classification of alcohols, with examples.34. Chemical properties of alcohols and phenols. 35. Identification of phenols.36. Identification of aldehyde group.37. Chemical properties of carboxylic acids.38. Monosaccharides: aldoses and ketoses; be able to draw structural formulas of an aldose and a ketose.39. Anomeric C atom.40. Mutarotation.41. Glycosidic bond in di- and polysacharide. Types of the bond.42. Omega series of fatty acids.43. Chemical properties of triacylglycerols. Acidic and alkaline hydrolysis: reaction equations with formulas of initial and

final reactants.44. Structure and properties of cholesterol.45. Classification of amino acids (know formulas of all 20 amino acids) 46. Zwitterions. Isoelectric point.47. Specific chemical reactions for amino acids and peptides (from lab works).48. Structures of nucleosides and nucleotides; their nomenclatures.49. Types of bonds in nucleosides and in nucleotides.50. Chemical structure of nucleic acids. Types of nucleic acids.

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MEDICAL CHEMISTRY. REQUIREMENTS FOR STUDY:

Practical work

1. The attendance of practical work is obligatory. Missing practical work is not allowed. 2. The practical work descriptions together with the report forms have to be printed out and bond into a

folder. 3. The practical work has to be defended during the time allotted for the particular laboratory

assignment. Students have to fill in the report form, complete additional tasks and prepare to answer questions provided in the description of the practical work. Only the defended practical work will be accepted.

4. Practical work will make 10 of the final assessment (up to 1 point).5. For the credit a student must have all practical work completed and defended.

Control tests

1. Students will take 2 control tests during the cycle on the material delivered during the lectures before the test.

2. The tests will be prepared as MCQs. To pass, a student needs to collect 40% of the correct answers. There will be one day to pass the test failed (only one test of the two by students choice).

3. The tests are obligatory.4. For the credit students have to pass at least 1 control test.5. The control tests will make 50 (1st – 25, 2nd – 25) of the final assessment (up to 5 points).

Final examination

1. The exam is a multiple-choice test of 50 questions.2. To pass, a student needs to collect 50% of the correct answers.3. The exam will make 40 of the final assessment (up to 4 points).

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SAFETY INSTRUCTIONS FOR WORK IN THE MEDICAL CHEMISTRY LAB

I. GENERAL REQUIREMENTS

1. A student is only allowed to work after listening to the safety instructions and signing in the registration book: a student must know the character of possible accidents and be able to provide first-aid.

2. A student must know:- Location of water and gas taps;- location of electrical switches;- location of fire extinguishers;- location of the first-aid kit;

and be able to use them properly in an emergency situation.

3. In the lab it is not allowed:- to be without a lab coat;- to disturb others by unnecessary talking and walking round;- to eat and drink;- to work in the absence of the teacher or technician;- to do things not related to the laboratory work;- to carry out experiments not included in the teaching plan;- to use damaged equipment and unfamiliar substances.

II. REQUIREMENTS BEFORE THE WORK

- to understand the task which is going to be carried out;- to know properties of the reagents to be used and properties of the reaction products, and to know how to handle them safely;- to know how to work properly and safely with the equipment and glassware required for that particular laboratory work;- the lab bench, on which you are going to do the experiment, must be clean and in order, without unnecessary equipment, glassware or personal belongings.

III. REQUIREMENTS DURING THE WORK

1. Avoid noisy behaviour, keep everything clean and tidy.

2. During laboratory work it is not allowed:- to leave working electrical equipment and burning gas burners unattended;- to leave the laboratory without teacher’s permission;- to work with damaged equipment or glassware;- to work with unknown substances;- to remove reagents from the places they have to stay;- to overload the lab bench with books, unnecessary reagents, glassware, etc.;- to leave gas burners and electrical equipment switched on when they are not needed;

3. Carefully follow the description of the laboratory work

4. When working with concentrated acids, bases, flammable substances or reagents with unpleasant odour, all the work must be done under a fume hood.

5. When heating over a flame, a test tube has to be inclined at an angle of 45 and pointed away from the working person and other students.

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6. Spilt reagents and broken glassware have to be cleaned immediately with appropriate precautions and under supervision of the technician.

IV. REQUIREMENTS AFTER THE WORK.

- turn off equipment, water and gas taps;- wash the glassware and clean the work surface;- place all the equipment, reagents and glassware in appropriate places.

V. FIRST-AID.1. First-aid to the injured person has to be immediate and correct. All people present in the room must be ready to help.2. If chemical compound gets on the face, eyes, hands or clothes, they have to be washed immediately with large amounts of water. 3. Every accident must be reported to the teacher.4. In the case of more serious injuries, intoxications or burns, an ambulance must be called immediately.5. In the case of electric shock, the power must be turned off immediately. If the victim is unconscious, cardiac massage and artificial respiration should be started right away.

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The example of the laboratory work report

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Results and calculations

Conclusion

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1. Preparation of solutions. Percent concentration

Background. Solution is a homogenous mixture containing two or more components. One of them, solvent, generally is present in the largest amount. All the remaining components are called solute and are equally distributed in the solvent. The physical state of solvent does not change. If solute is an ionic compound, it dissociates into ions when dissolved in the ionic solvent (water). The amount of solute per amount unit of solution (mass or volume unit) is called concentration, abbreviated as [C]:

[C] = amount of solute/amount unit of solventDepending on the units used to express the amount of solute and solvent there are several ways to indicate concentration. Thus, there is percentage concentration, molar concentration, molar concentration of equivalent and a few others. A widely used type of concentration is the percentage concentration. The term percent literally means number of parts in the total of one hundred parts. Consequently, the percentage concentration means number of solute parts in one hundred parts of solution. There are several types of the percentage concentration:

mass / mass – indicates number of solute mass units per 100 solution mass units and is denoted in parenthesis as (mass/mass), (m/m), (w/w). For example, 5 % NaCl (mass/mass) means that there are 5 grams of NaCl in 100 g of the solution;

volume / volume – indicates number of solute volume units per 100 solution volume units and is denoted in parenthesis as (vol./vol.) or (v/v). 10 % (vol./vol.) ethanol means that there are 10 ml of pure ethanol in 100 ml of the solution;

mass / volume – indicates the number of solute grams per 100 ml of solution. 3 % NaCl (mass/vol.) means 3 grams of NaCl in 100 ml of the solution. If the percentage concentration is shown without an indication in parenthesis, generally it means grams of solute per 100 g of solution. In clinical trials sometimes the obsolete milligram percentage concentration mg% is used, which indicates milligrams of solute in 100 grams of solution.

Experimental part: Prepare 250 g of CaCl2 solution in water of the following concentration: A 3%B 5%C 7%D 9%E 11%F 13%

1.Calculate the amount of salt and water needed.

Example: Prepare 250 g of 20 % CaCl2 solution in water. 20 % solution will contain 20 g CaCl2 in 100 g of solution;

x g should be in 250 g of solution; then x = 20g 250g/100g = 50 g CaCl2

250g – 50g = 200g H2OThus, 250 g of the solution will be made of 50 g of CaCl2, and the rest 200 g will be water.

2. Weigh CaCl2 on a piece of paper and transfer it to the flask.3. Since the density of water is approximately1g/cm3= 1g/mL, the weight of water will be equal to its volume. Use a graduated cylinder to measure the needed amount of water.4. Pour the water into the flask with the salt and dissolve the salt by mixing.5. Transfer the obtained solution into the cylinder and measure its density (d) with the aerometer. Find the real % concentration of this solution from the density tables. Indicate how much it differs from the theoretical value. Explain the possible error.

General questions

1. What systems are called solutions?2. What is a solvent and what is a solute?3. Describe the structure of water molecule in short.4. What is solution concentration?5. What types of concentration do you know?6. What is a percentage concentration? What types of percentage concentration are there?7. What does it mean: mass/mass, volume/volume, mass/volume concentration?8. How to determine the density of the solution and what density units are there?9. How can aerometer be used to determine the density of solution?10. What is the density of water and when is it the highest?

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Conclusion

2. Preparation of molar solutions: molar concentration

Background. Molar concentration - the method of expressing concentration that indicates how many moles of solute are present per unit volume of solution. The concentration of solution can be varied by using more or less solute or solvent, but in any case the molarity of solution is the number of moles of solute per liter of solution. The abbreviation for molarity is M. Molarity = number of moles of solute/ number of liters of solution,

Or using abbreviations: M = mol / LIt can be calculated dividing the amount of solute in moles (mol) by the volume of solution in liters (L):For example, if we have 4 moles of NaCl in 2 liters of solution, the molar concentration will be 4 mol / 2 L = 2 M (= mol/L). To remind you, one mole of a compound is numerically equal to the sum of atomic masses of all elements making up that particular compound. A concentration lower than 1M can be expressed using prefixes which mean:

milli- = 10-3; 1 mmol = 1 10-3 mol; 1mM = 1 10-3 M (mol/L)micro- = 10-6; 1 mol = 1 10-6 mol; 1M = 1 10-6 M (mol/L)nano- = 10-9; 1 nmol = 1 10-9 mol; 1nM = 1 10-9 M (mol/L)pico- = 10-12; 1 pmol = 1 10-12 mol; 1 pM = 1 10-12 M (mol/L)

Here are some examples of calculations using molar concentration:Example 1. What is the molarity of NaOH solution, if it contains 16.0 grams of NaOH in 2 liters of water solution?Molecular mass of NaOH = 23 + 16 + 1 = 40, thus, 1 mol of NaOH = 40 gNumber of moles of NaOH = grams of NaOH in solution / grams of 1 mol = 16 g / 40 g = 0.4 mol

Molarity of this solution = number of moles of solute / volume of solution == 0.4 mol / 2 L = 0.2 mol/L= 0.2 M

Example 2.24.5 mL of 1.5 M NaOH is added to the 20.5 mL of 0.85 M NaOH.What is the molarity of the final solution?It is equal to the amount of NaOH in moles in the initial portions of both solutions divided by the final volume of the mixture (in liters!)

moles of NaOH in the first solution:in (1 liter) 1000 mL there is -- 1.5 mol NaOHin 24.5 mL -- x mol

x = 24.5 mL 1.5 mol / 1000 mL = 0.0368 mol moles of NaOH in the second solution:

in 1000 mL -- 0.85 molin 20.5 mL -- x

x = 20.5 mL 0.85 mol / 1000 mL = 0.0174 molthe total number of moles of NaOH in both solutions = 0.0368 + 0.0174 = 0.0542 molthe final volume of the solution in liters = 24.5 mL + 20.5 mL = 45 mL = 0.045 Lthe molarity of NaOH in the final solution:

there is 0.0542 mol NaOH in 0.045 L of solution x -- in 1 L

x = 0.0542 mol 1L / 0.045 L = 1.2 M

The following example is of calculation is used in everyday lab work. Solutions usually are stored as stock-solutions of relatively high concentration (they are more stable), and working solutions are made from them by an appropriate dilution with water or other solvent. Such dilution is based on the fact that the number of moles of solute does not change during dilution:

moles of solute = molar concentration volume of solution; thus:Mbefore dilution Vbefore dilution = Mafter dilution Vafter dilution

Example 3.What volume of 1M KNO3 has to be diluted with water in order to get 250 mL of 0.2 M KNO3?

Put the numbers given into the equation above:1 M x = 0.2 M 0.250 L

then x = 0.2 M 0.250 L / 1 M = 0.05 L = 50 mL

Experimental: Prepare 250 ml of CaCl2 solution in water of following concentration:

A 0.5MB 0.8MC 0.9MD 1.3ME 1.5MF 1.2MFind the percent concentration (w/w and w/v) of this solution.

Procedure:1. Calculate the amount of the salt needed.

Example: Prepare 250 ml of 2M CaCl2 solution in water. Calculate its % (weight/volume) concentration.2.0 M CaCl2 means that in 1 liter (1000 ml) there are needed 2.0 moles of CaCl2;then in 250 ml there should be X moles of CaCl2

X = 250 2.0/ 1000 = 0.5 moles CaCl2 is needed for making 250 ml of 2.0M solution.

1 mol of CaCl2 amounts to 40 + 2×35.5 = 111 g;0.5 moles is X g

X = 0.5 111/ 1 = 55.5 gSo, to make the required solution we need 55.5 g of CaCl2

2. Weigh a salt on a piece of paper, and then transfer it into the flask.3. Use a graduated cylinder to measure approximately 150-200 ml of water.4. Pour the water into the flask; dissolve all the salt by stirring with a magnetic stirrer.5. Transfer the solution of salt back into the cylinder. Adjust the volume of the solution to 250 ml. Now you have the right amount of salt in the required volume of solution. However, the solution is not yet homogenous – you may notice disturbances and flows in the cylinder, indicating zones of various concentrations.6. Transfer the solution to the flask and mix it. The solution is ready.7. Calculate % (weight/volume) concentration of this solution:

Example: Solution has 55.5 g of salt in 250 ml X g in 100 mlX = 55.5 100/ 250 = 22.2 % (w/vol) concentration8. Determine the % (weight/weight) concentration:Transfer the solution into the graduated cylinder and measure its density with an aerometer. Find the real % (w/w) concentration to which the density of the obtained solution corresponds.

General questions

1. What is the mole?

2. How many elementary units does the mole contain?3. What kind of concentration is called molar concentration (molarity)?4. What is the dilution of solution?5. What principal works when solutions are diluted?

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Conclusion

3.Volumetric analysis. Acid-base titration

Background. In chemical analysis, it is often necessary to determine the concentration of ions in solution. For this purpose, a technique called titration is often used. It is based on the measurement of the volume of one reactant required to react with a measured mass or volume of another reactant in the chemical reaction. For this, a standard solution of a titrant (a reactant of known concentration) is added drop by drop to the measured volume of the solution of unknown concentration until the reaction is stoichiometrically complete. The task of titration is to determine what volume exactly of the titrant is needed to complete the reaction. The reaction is completed when all the particles of the substance have reacted with the particles of the titrant – this is called reaction endpoint or equivalence point. To make this point visible we use indicators. An acid-base indicator is a weak acid or a weak base. The undissociated form of the indicator has a different colour from its ionic form.

Ind-H <-------> H+ + Ind- (colour (1) <-----> different colour (2))

The colour change occurs over a range of hydrogen ion concentrations. This range is called colour change interval and is expressed as a pH range.Examples of the most widely used indicators:Methylorange -- red (pH 3.2) <<>> yellow (pH 4.4)Phenolphtalein -- colourless (pH 8.2) <<>> pink (pH 10.0)

Problem. Determine the amount of NaOH in solution of unknown concentration by titration with 0.1 M HCl.

ProcedureReaction which will take place NaOH + HCl ----> NaCl + H2OTo neutralize 1 mol of NaOH needs 1 mol of HCl

1. Fill the burette with 0.1 M HCl (titrant). Burette is a long tube, graduated in mL and tenths of mL, at the bottom it has a stopper which allows dripping of the solution. Mark the initial volume of the solution in the burette.2. Dilute the sample. Take a volumetric flask X with NaOH solution of unknown concentration (provided by the technician), and add distilled water up to the 100 mL mark. Close the flask and mix the contents thoroughly by turning over several times. 3. Pipette exactly 10 mL of diluted NaOH solution into a 100 ml Erlenmeyer flask; add 2 drops of phenolphthalein indicator. The solution colours pink.4. Put a magnetic bar in the flask and place it on the magnetic stirrer. While opening the stopper of the burette with one hand, slowly add HCl acid into the reaction flask. If titrating by hand swirl flask contents with another hand continuously. Near the endpoint slow the rate of addition to drops; the last few drops should be added at 3 -5 second intervals.5. Titrate until the colourless endpoint, which indicates that the neutralization reaction is complete. You need to catch the first moment of the colour change otherwise there would be too much titrant added. Record the final volume of the burette. The difference between the initial and final volumes is the volume of HCl required neutralizing all NaOH in the investigated sample.6. Repeat the titration once more. If the second measurement is close to the first, go to step 7, if not - titrate for the third time. Find the average of the closest two measuring.

7. Calculate the concentration of the given NaOH solution from the obtained titration data:-- reaction is completed when number of HCl (acid) moles added is equal to the number of

NaOH (base) moles in the flask:molesNaOH = molesHCl ; moles = Molarity Volume

Molarity NaOH VolumeNaOH = Molarity HCl VolumeHCl

Molarity NaOH = Molarity HCl VolumeHCl / VolumeNaOH

8. Calculate the amount of NaOH in the flask X in grams.

General questions

1. What is the method of volumetric analysis based on?2. What is the method of volumetric analysis (titration) used for?3. What solution is called “titrant”?4. What process is called “titration”?5. What is the equivalence point and what is its pH?6. What substances are acid-base indicators? How to choose a suitable one?7. What is the colour change interval of the indicator?8. At what pH does Methylorange change its colour?9. What colour does Phenolphtalein have when the pH is less than 8.2, and the pH is higher

than 10?10. What is the “titration curve”?

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Conclusion

4. Buffer solutions

Theory: Solutions of substances (usually in water) which can resist against changes of pH are called buffer solutions or buffer systems. The buffer solutions are made from a weak acid and its salt with a strong base (as the acetic acid/sodium acetate buffer) or a weak base and its salt with a strong acid (as

ammonium hydroxide/ammonium chloride buffer). As dissociation of weak acids or weak basis is extremely low, amounts of conjugated basis of conjugated acids are also low. Preparations of buffers solutions.

1. Selected weak acid is mixed with a salt containing common ion in the structure. This common ion acts as a base conjugated to selected acid. For example, acetic acid (CH3-COOH) and sodium acetate (CH3-COONa) have acetate (CH3-COO-) as a common ion. Wen acetic acid solution is mixed with sodium acetate solution, obtained buffer is known as acetate buffer.

2. Salts of different acidities can also be used for preparation of buffers. For example, NaH2PO4 (acts as an acid) and Na2HPO4 (acts as a base), which dissociation results in ions of different acidities:NaH2PO4 Na+ + H2PO4

- Na2HPO4 2Na+ + HPO4

2- (base)H2PO4

- ion contains 2H+ ions and acts as a weak acid, HPO42- ion has a single H+ therefore it acts

as a base conjugated to H2PO4-.

pH of buffer solutions.

Let consider a case when buffer comes from an acid HA and a salt of the acid MeA. In water, an acid undergoes ionisation by this equation (for simplicity take dissociation):HA H+ +A-

Constant of acidity for such dissociation is described by an equation:

Concentration of protons (H+ ions) can be calculated from re-arranged equation:

If we add salt in the solution of such acid, dissociation of the acid will be depressed by increased amounts of the ion A-. It means, that concentration of non-dissociated acid [HA] will be equal to concentration of the acid added at the beginning. Therefore H+ concentration in such acidic buffer can be calculated by the formula:

We can calculate pH of such a buffer solution using Henderson-Hasselbalch equation:

We can also prepare a buffer solution if we mix different volumes of solutions of an acid and a base. In this case Henderson-Hasselbalch equation is:

In Henderson-Hasselbalch equation, Vsalt is a volume of salt solution Vacid is a volume of acid solution [salt] is a concentration of salt solution taken for preparation of a buffer [acid] is a concentration of acid solution taken for preparation of a

buffer

Ka= [H+]x[A-][HA]Ka= [H+]x[A-][HA]

H+ = Ka x [A-][HA]

H+ = Ka x [A-][HA]

H+ = Ka x [salt][acid]

H+ = Ka x [salt][acid]

pH= -lg[H+]= lg [salt][acid]

- lgKapH= -lg[H+]= lg [salt][acid]

- lgKa

pH= -lg[H+]= lg [salt] x Vsalt[acid] x Vacid

- lgKapH= -lg[H+]= lg [salt] x Vsalt[acid] x Vacid

- lgKa

Mechanism of buffer action. H+ concentration and pH of buffer solution do not change (or change very little) when acid or base is added to them. This phenomenon comes from interaction of buffer components with added acid or base. Suppose we add HCl (strong hydrochloric acid) into acetate buffer (composition was considered above):

CH3-COONa + HCl CH3-COOH + NaClCH3-COO- + H+ CH3-COOH

As CH3-COOH (acetic acid) undergoes very low dissociation, pH is little changed. Little change of pH comes from alterations in the ratio between salt and acid amounts.If a strong base is added to acetate buffer, it combines with acidic component of the buffer:

CH3-COOH + NaOH CH3-COONa + H2OCH3-COOH + OH- CH3-COO-+ H2O

This example demonstrates increasing in salt amount in the buffer solution.

Buffer capacity. A buffer solution can keep pH stable only up to a certain amount of acid or base added. After reaching this threshold pH of a buffer solution changes if extra acid/base is added, as it does in the case of regular solutions. Therefore, every buffer solution is characterized by the buffer capacity (B) which indicates how many moles of a strong acid or base should be added to 1 litre of the buffer in order to change its pH by 1 pH unit: B = C / (pHafter addition of acid or base - pHinitial); where B - buffer capacity, C - acid or base concentration mol/L The buffer capacity depends on the nature and concentration of the buffer components as well as on the ratio of these concentrations:

- The buffer capacity increases when the concentration of buffer components increases;- The capacity for both acid and base is highest in a buffer where ratio of concentrations equals 1.

ExperimentalPrepare acetate/sodium acetate buffer. Calculate its pH and buffer capacity. Each pair of students has to prepare and analyse only one buffer solution. Procedure

1. Take 2 flasks and, mixing the appropriate volumes of two components -- 0.1 M acetic acid and 0.1 M sodium acetate in each flask prepare 2 identical acetic buffers (one - for determining buffer capacity for acid, and the other - buffer capacity for base). The volumes are indicated in the table. Draw the table in your notebook and fill it in with the data of your own experiment and the experiments of other students:

Vbuffer=VCH3COOH

+VCH3COO-Na

=10 ml

pH pH V (titrant) Buffer capacity

N

CH3COOH,mL

CH3COONa,mL

InitialpHi

After titration with acid

After titration with base

pHacid pHbase Vacid Vbase Bacid Bbase

123456

234567

876543

3.43.43.43.43.43.4

6.36.36.36.36.36.3

2. Calculate the initial pH of the prepared buffer using Henderson-Hasselbalch equation:

pH = lg ([salt] / [acid]) - lg Ka; Ka - ionization constant for the acetic acid, Ka= 1.85 10-5, and pKa=-lg (1.85 10-5) = 4.7.Since the concentrations of components in the buffer are equal, the volumes in mL (used in preparation of the buffer) can be used instead of the concentration:

pH = lg ( volume of salt / volume of acid ) + 4.72. Add 3--4 drops of methyl-orange indicator to one flask. Titrate this flask with 0.1 M HCl until the

pink-orange colour appears (indicating pH = 3.4 – that is pH after titration with acid). Use the titration volume of HCl (Vacid) to calculate the buffer capacity for acid Bacid:

Bacid = (Vacid Cacid / pHacid Vbuffer)Vbuffer - volume of buffer in ml,pHacid - change of pH after titration with acid.

3. Add 3-4 drops of methyl-red indicator to the second flask. Titrate this flask with 0.1 M NaOH until the yellow colour starts to appear (pH = 6.3 – the pH after titration with base). Use the titration volume of NaOH (Vbase) to calculate the buffer capacity for base Bbase:

Bbase = (Vbase Cbase / pHbase Vbuffer) pHbase -- change of pH after titration with base.5. Copy the results of other students who were analysing different buffers.6. Draw the graphs of Bacid and Bbase dependence on pH initial. The best buffer capacity for both acid and base you should get around pKa value (pH=4.7).

General questions

1. What solutions are called “buffers”?2. How are buffer solutions prepared?3. What does the pH of buffer solutions depend on?4. What equation is used to calculate the buffer pH?5. How does acetic acid/acetate buffer maintain solution pH at almost the same level if a small

amount of strong acid is added? Write the equations.6. How does acetic acid/acetate buffer maintain solution pH at almost the same level if a small

amount of strong base is added? Write the equations.7. What is called the buffer capacity?8. How can buffer capacity for acid be calculated? Write the formula.9. How can buffer capacity for base be calculated? Write the formula.10. What does buffer capacity depend on?11. Why do biological fluids have a stable pH?

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Conclusion

5. Colloidal solutions

Background. Colloidal solutions are disperse systems where disperse phase particles are 1-100 nm in size. Such solutions are stable over time. The solution where disperse phase is solid and medium is liquid is called sol. To prepare a colloidal solution, the particles of disperse phase have to be made of the size of colloidal particles. There are two ways to do this: condensation (to make the colloidal size particles from smaller ones) and dispersion (to disperse larger particles into colloidal size particles).

Dispersion methods:

- Colloidal mill (a mechanical way to grind large particles);- Ultrasound;- Peptization (disaggregates large particles by chemical agents, which increase repulsion of

particles).Condensation methods (produce compounds of relatively low solubility from soluble ones):- Exchange of solvent (by another solvent in which the same substance has lower solubility);

-Oxidation (obtains neutral chemical elements (mainly non-metals) from their ionic forms);- Reduction (obtains neutral elements (mainly metals) from their ionic forms);- Hydrolysis reaction (makes compounds of lower solubility);

-Exchange reactions (makes insoluble compounds).

Exchange reaction: AgNO3(aq) + KCl(aq) → AgCl↓(s) + KNO3(aq). Micelle when it is excess of AgNO3: {m(AgCl)nAg+(n-x)NO3

-}x+NO3- .

Micelle when it is excess of KCl : {m(AgCl) nCl-(n-x)K+}x-K+.

ExperimentalA. Prepare colloidal Fe(OH)3 solution (sol) by chemical condensation (hydrolysis reaction):

1. Put 1 mL of 2% FeCl3 into a test tube, add 10 mL of distilled water.2. Mix thoroughly and boil the mixture until brown transparent Fe(OH)3 sol. is formed. 3. Write the hydrolysis reaction and micelle of Fe(OH)3.

Reaction of hydrolysis: FeCl3 + H2O→……………………………………………………………..…………………................................................................................................................................................................................................................................................................................................Micelle: {m(Fe(OH)3●●●●●. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

B. Preparation of colloidal colophony (pine resin) sol by the exchange of solvent1. Add several drops of 2% colophony/ ethanol solution to 10 mL of distilled water. 2. Mix the mixture. Milky sol should be obtained.3. Explain, how and why this sol was formed.

C. Stability and coagulation of colloidal solutions.The coagulation of colloidal solutions usually occurs when an electrolyte solution is added to it. The coagulation means that the particles became larger (more than 100 nm), and these newly formed particles easily sediment. Prepare 5 test tubes for each electrolyte (one pair of students experiments with one electrolyte), and fill them accordingly to the table:

Number of the test tube 1 2 3 4 5Fe(OH)3 sol (prepared by the technician), mL 5 5 5 5 5H2O, mL 4.5 4 3 2 1Electrolyte, mL

1) 3 M KCl2) 0.01 M K2SO4

3) 0.001 M K3[Fe(CN)6]

0.50.50.5

111

222

333

444

Leave the test tubes for 30 min, and then observe where the colloidal solution has coagulated. The test tube with the lowest concentration of electrolyte where coagulation was observed has concentration called coagulation threshold. Coagulation threshold is the lowest electrolyte concentration (mmol/L) which induces coagulation of the sol. Calculate the coagulation thresholds:

)/()/(

LmmolVVVxVLmmolC

Cwatersoleelectrolyt

eelectrolyteelectrolytthr

The opposite number of the coagulation threshold is called the coagulation power:thrC

P 1 .

Write the results in the table given below. Discuss the results with other students who have done the experiment with different electrolytes. Make a conclusion of which electrolyte is a better coagulation agent and explain why.

Electrolyte Coagulating ion Coagulation threshold Coagulation power

KCl

K2SO4

K3[Fe(CN)6]

General questions

1. What is called the disperse phase and the dispersion medium?2. What solutions are called “colloidal solutions”?3. What systems are called “sol”?4. Provide some examples of disperse systems in living organisms.5. What properties should chemical substances have, so that they can be used to prepare

colloidal solutions?6. What methods of colloid preparation are there?7. What method of preparation of colloids is called peptization?8. What is micelle and what is it made of?9. What is called coagulation, coagulation threshold and coagulation power?10. How can electrolytes cause coagulation?

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Conclusion

11.

6. High molecular mass compounds and their properties.Swelling and gelatinization of high-molecular mass compounds

Background. High-molecular mass compounds are organic compounds with molecular mass as big as hundreds of thousands or even millions of Daltons. By structural features, they usually are polymers two origins:

1. Cellular origin high-molecular mass compounds are called biopolymers, e.g. nucleic acids, proteins and polysaccharides.

2. Synthetic/artificial high-molecular mass compounds.

Sizes of particles of high molecular mass compounds are similar to colloidal particles, but they are water-soluble. Therefore solutions of high molecular mass compounds share some common properties with colloidal solutions and with real solutions. They also have very specific properties. Properties of solutions of biopolymers:1. Properties that are common with colloidal solutions: Size of molecules of biomolecules is similar to the size of colloidal particles. Solutions of biopolymers have slow rate of diffusion. Molecules of biopolymers cannot pass through semi-permeable membrane, what implies that their

solutions have low osmotic pressure. The osmotic pressure of those solutions depends only on number of biopolymer molecules.

2. Properties of biopolymer solutions, that are common with those of real solutions: Do not form micelles, i.e. they are solutions of a single phase (homogeneous systems). Solutions of biopolymers with linear structure do not show Tindall effect. Solutions of biopolymers are stable systems. They do not show sedimentation phenomenon.3. Specific properties of biopolymer solutions. Biopolymers swell before dissolving. Under swelling solvent surrounds the molecule of biopolymer,

then it moves into the empty spaces of molecule, so that the volume of the molecule increases.

Swelling of high-molecular mass compounds. Swelling of a high-molecular mass compound results in increasing of both mass and volume of a high-molecular mass compound. There are two sorts of swelling: the limited swelling and unlimited swelling where a solution of high-molecular mass compounds is formed. Swelling depends on the properties of the solvent.

The swelling degree (Q) is the main characteristic of swelling. It shows how much the volume of the high molecular mass compound enlarges during the swelling. The swelling degree shows how many cm3 of the solvent can be absorbed by 1 cm3 of a high-molecular mass compound:

V1 is the volume of a high-molecular mass compound before swelling,V2 is the volume of a high-molecular mass compound after swelling.The swelling degree (Q) can be estimated by weighing of a high molecular mass compound before its swelling and after swelling. The swelling is affected by:

1. Temperature. As swelling is an exothermic process, increasing in temperature results in decreasing of the swelling degree.

2. Pressure. Increasing in pressure results in increasing of the swelling degree.3. Electrolytes and pH of solution. Swelling is little at isoelectric point, at which molecules can stick

together and entering of water is slow.4. Degree of dispersion. High dispersion degree favours to swelling. 5. Time.

Jellification (gelatinization). Solutions of high-molecular mass compounds can form a jelly. It is a net-like structure formed in the solution when hydrophobic regions of the molecules interact with each other and the hydrophilic regions are highly hydrated, water molecules filling in the gaps between the molecules. In chemistry, the jelly is called gel. The process when solutions of high-molecular mass compounds lose their fluidity is called jellification. It depends on the following factors:

V2 – V1 V1

Q =

1. Concentration of high-molecular mass compound solutions. Jellification can occur in concentrated solutions of high-molecular mass compounds.

2. Size and shape of molecules of a high-molecular mass compound. Thread-like molecules can easily form gel, but the ball-like ones can hardly do this.

3. Temperature. Low temperatures favour to jellification.4. Presence of electrolytes and pH. Anions are the most important for the jellification. According to the

efficiency of the effect, the anions make a line:SO4

2- >citrate>CH3-COO- >Cl- >NO3- ->Br- >I- >SCN-

Jellification is promoted by highly hydrated ions. Starting with Cl - the anions diminish the jellification (gelatinization). These ions are adsorbed on the surface of the macromolecules. They give charge to polymers and prevent macromolecules from jellification.

Lab. Procedure

Effect of solvent on swelling.

1. Take 4 test tubes. Put 2ml of water in the tubes N1 and N2. Put 2 ml of benzene into the tubes N3 and N4.2. Add one piece of agar (polysaccharide) into the tube N1 and another piece of the same size into the tube N3.3. Add one piece of synthetic rubber into the tube N2 and another piece of the same size into the tube N4.4. After 20 min., compare the sizes of the agar and rubber. Make a conclusion about the effect of solvents on the swelling.

Effect of electrolytes on jellification (gelatinization)

1. Take 6 test tubes and put 1.5 ml of 1 M solutions of the electrolytes as indicated in the table:Number of test tube 1 2 3 4 5 6

Electrolyte K2SO4 CH3COOK KCl KI KSCN H2O

Beginning of gelatinization t1

End of gelatinization t2

Time required for complete gelatinization t2 - t1

2. Add 1.5 ml of 6% hot gelatine solution into each test tube and mix thoroughly. 3. Place the test tubes into the hot water bath (50 - 60º C) for 10 minutes.4. Remove the test tubes from the bath and place them in cold water. Mark the time t1.5. Periodically check the fluidity of the solution by inclining the test tubes. If you observe that solution is no longer fluid, mark the end time of gelatinization.6. Fill in the table. Make a conclusion about which ions favour gelatinization and which do not.

General questions

1. What substances are called the high molecular mass compounds (HMMC)? Give some examples.

2. How are the solutions of HMMC similar to the colloidal solutions?3. How are the solutions of HMMC similar to the real solutions?4. What properties are specific to the high molecular mass compounds?5. What process is called swelling?

6. What effect on the swelling does the solvent have?7. What does the swelling degree show?8. What process is called the jellification?9. How does the gel form and what can affect its formation?10. What effect do the electrolytes have on jellification?

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Conclusion

7. Coordination compound based determination of water hardness

Background. Some amino-polycarbonic acids and their salts can form stable coordination compounds with most of the cations. These compounds are called complexones (chelating agents). An example of such complexones is EDTA (ethylendiaminetetraacetic acid):

Most often the disodium salt of EDTA is used, which has a trivial name of Trilon B. This compound reacts with the divalent ions (such as Ca2+, Mg2+, Ba2+) forming a rather stable colourless coordination compound. One molecule of Trilon B binds one ion:

Na2[H2EDTA] + Ca2+ <-----> Na2[CaEDTA] + 2H + [H2EDTA] 2- + Ca2+ <-----> [CaEDTA] 2- + 2H +

The term “total water hardness” means a total amount of Ca2+ and Mg2+ ions dissolved in water. Due to the ability to bind these ions and to form a coordination compound Trilon B is used for the quantitative determination of these ions measuring the water hardness. Since the compound of Trilon B with Ca2+ and Mg2+ ions is colourless, an indicator is needed to pinpoint the equivalence point. The most suitable for this purpose is Eriochrom Black T (for convenience abbreviated as H3Ind). Depending on pH this indicator has different colours: pH 6 – red; pH 7-11 – blue; pH 12 – yellow-orange

H3Ind ↔ H2Ind- + H+; pH = 6

red

H2Ind- ↔ HInd2- + H+; pH = 7-11

blue

HInd2- ↔ Ind3- + H+; pH = 12

yellow-orange

Eriochrom Black T forms a red-violet coordination compound with Ca2+ and Mg2+ (at pH>7), which is less stable than the one formed by Trilon B. While titrated with Trilon B, all Ca2+ and Mg2+ ions from the solution are coordinated by it, then this red-violet compound decomposes and more stable compound of Ca2+ and Mg2+ with Trilon B forms. The free anions of Eriochrom Black T make solution blue-coloured at the equivalence point:

[MgInd] - + [H2EDTA]2- ------> [MgEDTA]2- + HInd2- + H+

Red-violet colourless colourless blue

Water hardness values:

< 0.75 mmol/L – soft water1.5 – 2.7 mmol/L – medium soft/hard water2.7 – 5.35 mmol/L – hard water> 5.35 mmol/L – very hard waterExamples: rainwater – 0.05 mmol/ L; sea water – 65 mmol/ L.

ExperimentalProcedure

1. Dilute the given solution X (solution of MgCl2 whose hardness has to be determined) to the 100 ml in a volumetric flask (add distilled water up to 100 ml)2. Put 10 mL of this diluted solution X into the flask, add 2 mL of the buffer solution (1 M NH4Cl/NH4OH) and 4 drops of the indicator –Eriochrom Black T. Mix well.3. Titrate the mixture with the solution of 0.05 M Trilon B (mix well during titration) until the red colour starts changing into violet (blue). 4. The titration is over when after one single drop of Trilon B the solution becomes blue (You need to catch the first moment of the change in the initial colour). Therefore the last portion of Trilon B should be added drop-by-drop. Record the volume (mL) of Trilon B used for titration.5. Repeat the titration three times, and take the mean volume of the three titrations.6. Calculate the total water hardness. One mole of Trilon B can bind one mole of Ca2+ or/and Mg2+ ions. Use the equation:

HT = (MB VB / Vwater ) 1000 mmol/L,where HT -- total water hardness in mmol/L; MB – molarity of the Trilon B solution; VB – volume of the Trilon B used for titration; Vwater – volume of water used for titration.

7. Calculate how many milligrams of MgCl2 are in your X solution (in 100 ml of solution).

General questions

1. What compounds are called coordination compounds? Give some examples.2. What compounds are called complexones (chelating agents)? Give some examples.3. Draw the formulas of EDTA and disodium salt of EDTA.4. What is the trivial name of the disodium salt of EDTA?5. What does the term “total water hardness” mean?6. Why is the indicator Eriochrom Black T used when total water hardness is determined? 7. Write down the equation of the reaction between Eriochrom Black T, Trilon B and metal ions.

Explain why the colour of the solution changes at the equivalence point.8. Write down the formula which is used to calculate the total water hardness.

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9. Chemical kinetics

Experimental. A. Rate of chemical reaction determined by change of reaction product concentration.This is the reaction of sodium thiosulphate decomposition:

Na2S2O3 + H2SO4 -------> Na2SO4 + S + SO2 + H2OOne of the products is insoluble in water – it is elementary sulphur S. The increased turbidity of the reaction mixture provides a pretty good way to evaluate the reaction rate. The increased turbidity means that more sulphur was formed, and the sooner this process occurs the higher is the rate of the above mentioned reaction. Procedure

1. Take a small beaker and fill it according to the table (draw it in your notebook) as N 1. As it is obvious from the table’s second column, the following flasks will have increasing sodium thiosulphate concentrations, as the total volume in each of them will be the same:

Number 0.25 M Na2S2O3; mL

H2O,mL

Time,s (seconds)

Rate of reaction,1/s

N 1N 2N 3

246

420

2. Add 2 mL of 0.25 M H2SO4, (use a little graduated cylinder for that, not a pipette), stir carefully with a glass rod. Start time countdown from the moment of adding the acid. Immediately put this flask on the piece of paper lined in squares. While slowly stirring the contents of the flask, observe when the lines first become obscured by the layer of reaction mixture. Record the time elapsed from the moment of H2SO4 addition. Calculate the reaction rate.

3. Wash the beaker carefully and repeat the same procedure with the other concentrations of sodium thiosulphate N 2 and N 3.

4. Draw the graph of the dependence of the time needed for the reaction (in seconds) on the volume of substrate added (in mL), (mL should be on the X axis, and seconds -on the Y axis).

B. Effect of catalyst on the reaction rate Catalyst is the substance which changes the reaction rate when present in the reaction mixture,

but remains unaltered itself. The reaction rate changes because catalyst decreases the activation energy of that particular reaction. This is the reaction of hydrogen peroxide decomposition:

2 H2O2 ------> 2 H2O + 2 [O], then2[O] -------> O2

The catalyst for this reaction is MnO2. In living organisms this reaction is catalysed by the enzyme catalase. Decomposition of hydrogen peroxide at the room temperature occurs very slowly, and the bubbles of formed oxygen are not visible. When MnO2 is added, the reaction rate increases significantly, and this can be noticed by vigorous bubbling.ProcedurePut 1 mL of 3% H2O2 solution into the test tube. There are no bubbles. Add a little of MnO2 powder and cover the test tube with your finger while the reaction proceeds. At that time get a wooden splint burning and blow the flame out. Stick a glowing wooden splint inside the test tube. Observe what happens and explain the result.

C. Equilibrium shift.

A chemical reaction consists of direct and reverse reactions, going on in opposite directions. The term “Equilibrium of chemical reaction” means that equilibrium between the direct and reverse reaction has been reached, and the rate of direct reaction is equal to the rate of reverse reaction. Equilibrium of the chemical reaction can be disturbed by variety of factors among which is change of: 1) Concentration of reactants or reaction products; 2) Temperature; 3) Pressure; 4) Volume of the system. In living organisms the main factor affecting equilibrium of chemical reaction is the change of substrate or product concentration, since temperature, pressure and volume are kept constant. The effect of various factors on the equilibrium of chemical reaction is described by Le Chatelier’s principle:

If a system in equilibrium has been disturbed by the outside factors (as concentration, temperature or pressure) then the equilibrium of this system shifts in the direction which tends to decrease the effect of outside factors.

I.

CoCl2 dissociates when dissolved in water:CoCl2 ↔ Co2+ + 2Cl-

Then the ions formed make a coordination compound where Co (II) has additional bonds with water: Co2+ + 2 Cl- + 6 H2O ↔ [Co (H2O)6]Cl 2 ↔ 2 Cl- + [Co(H2O)6]2+ (red colour).

When HCl is added, the red-coloured coordination compound transforms into another blue-coloured:(Red colour) [Co (H2O)6]2+ + 2 Cl- + 2 Cl- ↔ 6 H2O + [CoCl4]2- (blue colour)

Keq = [[CoCl4] 2-] [H2O]6 /[ [Co(H2O)] 2+] [Cl-]4

When HCl is added, the equilibrium of the reaction shifts towards formation of [CoCl4]2-. If extra water is added the equilibrium shifts in the opposite direction.

Procedure1. Dissolve a crystal of CoCl2 in a test tube with few drops of water.2. Add several drops of concentrated HCl until the solution becomes blue.3. Dilute this solution with water until reappearance of less intense red colour.4. By heating and cooling the test tube find what is the effect of temperature on this equilibrium.

II.

FeCl3 reacts with NH4SCN and red solution of the iron rhodanide forms:

FeCl3 + 3NH4SCN ↔ Fe(SCN)3+ 3NH4Cl.

This reaction is a reversible one. The shift of the equilibrium could be seen by observing the intensity of the red colour in the solution. The colour intensity depends on the concentration of the iron rhodanide – Fe(SCN)3. When the concentration of the Fe(SCN)3 in the solution is high, the red colour of solution is intense. When the concentration of the Fe(SCN)3 is low the solution is faded red. It is possible to shift the equilibrium by adding more reactants or products.

Procedure

1. Pour into a beaker 20 mL of water and add 3 drops of the saturated solution of the FeCl3 and 3 drops of the saturated solution of the NH4SCN.

2. Divide the solution into four test tubes.3. Add 2 drops of the FeCl3 into the first test tube.4. Add 2 drops of the NH4SCN into the second test tube.5. Add some crystals of the NH4C l into the third test tube.6. Stir the tubes and compare colour intensity in these test tubes.7. Describe and explain the observed differences.

Calculation of Michaelis-Menten constant and maximal reaction rate

Michaelis-Menten equation describes the relationship between reaction rate and substrate concentration in enzyme-catalysed reaction.

[S]K[S]VV

m

max0

Vmax is the maximum velocity of the reaction – when all enzyme molecules are fully active. Km – Michaelis-Menten constant – is the concentration of substrate at Vmax/2.Task: calculate Michaelis-Menten constant and maximal reaction rate for the given enzyme-catalysed reactions.Procedure:1. Calculate reaction rates (v) with different substrate concentrations. time reaction

formed product of amountv mol/min.

2. Plot 1/[S] on x axis and 1/v – on y axis.3. Find the Km and Vmax values for the given reactions from Lineweaver-Burk plot.

Version A. Assay conditions were following:1 mg of the enzyme was incubated with the indicated concentration of substrate [S] for 5 minutes, after which the

amount of formed product was determined and is presented in the table:Substrate concentration (μM) 0.63 0.83 1.25 3.3

Amount of product formed per 5 min (μmol) 29 36.2 47.2 76.9

Version B. Assay conditions were following:1 mg of the enzyme was incubated with indicated concentration of substrate [S] for 5 minutes,

after which the amount of formed product was determined and is presented in the table:Substrate concentration (μM) 0.67 0.91 1.4 2.5

Amount of product formed per 5 min (μmol) 37 43.5 53 62

Version C. Assay conditions were following:1 mg of the enzyme was incubated with indicated concentration of substrate [S] for 5 minutes, after

which the amount of formed product was determined and is presented in the table:Substrate concentration (μM) 0.42 0.71 1.25 2.5

Amount of product formed per 5 min (μmol) 116 147 200 250

Calculation of Km and Vmax values

General questions

1. What does chemical kinetics study?2. How is the rate of chemical reaction expressed?

Lineweaver-Burk Plot (1)

-2.0 -1.5 -1.0 -0.5 0.5 1.0 1.5 2.0

-0.1

0.1

0.2

0.3

0.4

0.5

1/[S]

1/v


-2.0 -1.5 -1.0 -0.5 0.5 1.0 1.5 2.0

-0.1

0.1

0.2

0.3

0.4

0.5

1/[S]

1/v


-3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.5 1.0 1.5 2.0 2.5 3.0

-0.1

0.1

0.2

0.3

0.4

0.5

1/[S]

1/v

3. What factors can make an impact on the rate of chemical reaction?4. How does the reaction rate depend on the concentrations of reactants?5. How is it possible to determine the reaction rate of this reaction?

Na2S2O3 + H2SO4---> Na2SO4 + S+ SO2 + H2O

6. What does a catalyst do when it is present in a reaction mixture? How does it work?7. Which catalyst can catalyze the reaction of hydrogen peroxide decomposition?8. What does “the equilibrium of chemical reaction” mean?9. What factors can influence the equilibrium of chemical reaction? 10. Describe “the Le Chatelier’s principle”.11. Why and how can the CoCl2 solution change its colour?12. What factor determines the intensity of colour of iron rhodonide solution?13. What does the Michaelis-Menten equation describe? Draw this equation.14. What does „the Lineweaver-Burk plot“ mean? Draw this plot.15. How could the Michaelis-Menten constant and maximal reaction rate for the given enzyme-

catalysed reaction be calculated?

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Conclusion

10. Chemical properties of carbonyl compoundsEsters and lipids

Carbonyl compounds are organic compounds containing carbonyl (oxo) group in a molecule. The compounds fall into two big groups – aldehydes and ketones. In aldehydes, the carbonyl group has both

hydrogen atom and hydrocarbon radical attached. In ketones, the group is bound with two hydrocarbon radicals. According to hydrocarbon radical present, aldehydes are grouped into aliphatic ones (a), alicyclic ones (b), and aromatic ones (c):

Ketones are also grouped into aliphatic (a), alicyclic, and aromatic.

Aldehydes and ketones with up to 4 C atoms in molecules are volatile liquids of specific odour. They are soluble in water and in organic solvents. Solubility in water decreases with an increase of a number of C atoms in a chain. Aldehydes containing 8-10 C atoms in a chain have odour of flowers and are used in perfumery.

Chemical properties of carbonyl compounds

Properties of carbonyl compounds are determined by chemical properties of both the carbonyl group and hydrocarbon radical.1. Reduction producing alcohols (both aldehydes and ketones):

2. Oxidation producing acids (only aldehydes):

In this laboratory work you will prove chemical properties of carbonyl compounds.

Experimental

A. Oxidation of aldehydes (12) Free aldehyde groups can be oxidized to carboxyl easily by mild oxidizing agents – for example, Cu++

in alkaline solution:

H3C CO

H

CO

HC

O

H

a b cethanal cyclohexanecarbaldehyde benzaldehyde

H3C C CH3

O

a

propanone(acetone)

H3C C CH3

O

propanone(acetone)

H3C C CH3

OH

HH2

H-C-H H-C-OH

= =

O O

Formaldehyde Formic acid

R-CHO + 2 Cu++ + 5 OH- + heating R-COO- + Cu2O + 3H2O Blue solution Red precipitate

Trommer’s reagent: mix of CuSO4 and NaOH;Fehling’s reagent: copper is coordinated with sodium potassium tartrate (Fehling-I – CuSO4, Fehling-II – basic sodium potassium tartrate)

Procedure 1: Oxidation by Trommer’s reagent1. Take 2 test tubes and add 6 drops of 2 M NaOH, 6 drops of H2O and 2 drops of 0.1 M CuSO4

solutions into each tube. The blue Cu (OH)2 precipitate should form in them.2. Add a few drops of formalin (the solution of formaldehyde) to the 1st tube and a few drops of acetone to the 2nd tube and shake.3. Heat the upper part of the solution in the tubes over the flame until the reaction mixture changes colour from blue (Cu++) to the yellow (CuOH) or red (Cu2O).4. Describe and explain the observed differences.

Procedure 2: oxidation by Fehling’s reagent1. Take 2 test tubes; put 5 drops of formalin into the 1st tube and acetone into the 2nd tube.2. Add 10 drops of Fehling-I and 10 drops of Fehling-II reagents into each tube.3. Heat the upper part of the solution in the tubes over the flame until the colours of reduced copper compounds appear.4. Describe and explain the observed differences.

B. Reaction of acetone with nitroprusside Sodium nitroprusside is a coordination compound Na2[Fe(CN)5NO] which forms with acetone the complex compound of intense red colour:

CH3-CO-CH3 + Na2[Fe(CN)5NO] + 2NaOH → Na4[Fe(CN)5NO=CH-CO-CH3] + 2H2O

Procedure: 1. Add 1 drop of 0.25M sodium nitroprusside solution, 5 drops of water and 1 drop of acetone in the test tube.2. Add 1 drop of 2M NaOH – solution becomes red. Add 1 drop of 2M CH3COOH – the colour becomes more intense – almost cherry.

C. Determination of products of lipid peroxidation Plant oils, containing high amount of unsaturated fatty acids, during prolonged storage in the light place at room temperature can gain bitter taste and unpleasant odour. It happens due to lipid peroxidation. Peroxidation involves free radical formation and results in the fragmentation of unsaturated chain, forming products with shorter carbon chain – aldehydes and acids. Reactions for determination of peroxides in oil:

[O] (from peroxides in oil) +2KI +2CH3COOH 2CH3COOK +H2O +I2

I2 + starch blue complex compound

The reaction of the formed iodine with sodium thiosulfate can be used to determine the degree of lipid peroxidation in oil:I2 + 2Na2S2O3 2NaI +Na2S4O6

Procedure: 1. Take 2 test tubes. Add 2 drops of fresh oil into the 1st tube and add 2 drops of old rancid oil into the 2nd tube.2. Add 10 drops of acetic acid and chloroform mixture to the each tube, then add 5 drops of 0.5 M KI to the each tube and shake.3. Add 2 drops of 0.5 starch solution into each tube. Observe the colour change.

4. Take the tube with the blue solution and drop-by-drop (count the drops added) add 0.05 M sodium thiosulfate until the blue solution becomes colourless. Record the number of thiosulfate drops required to get the colour change.

General questions

1. What compounds are called “carbonyl compounds”? What groups of them are there?2. What chemical properties do carbonyl compounds have?3. Why aldehydes can react with copper sulfate, and ketones cannot?4. What principle of chemical reaction works between aldehyde and copper sulfate (CuSO4)?5. Draw the equation of reaction between formaldehyde and CuSO4. Explain why the colour of the

solution changes.6. What chemical substances occurs compose Fehling’s reagent?7. What type of reaction does occur when acetone reacts with Na2[Fe(CN)5NO]?

8. What processes do take place in plant oils during prolonged storage in a light place at room temperature? 9. What reactions are used for the identification of peroxides in oil?

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Conclusion

11. Chemical properties of alcohols, phenols and amines

Background. Alcohols are chemical substances which contain one or more hydroxyl group (-OH). The possession of this group provides some specific physical and chemical properties. Alcohol can form

hydrogen bonds. As a result of this they are soluble in water. The hydroxyl group gives some acidic properties. Alcohols can react with active metals: 2CH3-CH2-OH + 2Na → 2CH3-CH2-ONa + H2

When alcohol contains more than one -OH group it can form blue coloured coordination compound with the basic Cu(OH) 2:

CH2

CH2

OH

OH+ Cu(OH)2 + Cu

O

O

O

O

CH2

CH2

CH2

CH2

2K+ + 4H2O

blue colourprecipitate

water solubleblue complex compound

KOH2

2-

The oxidation of primary alcohols results in the formation of aldehydes (1), and then aldehydes undergo

further oxidation to carboxylic acids.

R CH2 OH

[O]-H2O

R CO

H

[O]R C

O

OH

Under the oxidation of secondary alcohols, ketones are produced (2). Ketones are more resistant to

oxidation.

R CH R1

OH

[O]-H2O

R C R1

O

Alcohols can be oxidized using strong oxidizers. Mild oxidizers as Trommer’s and Fehling’s reagents cannot oxidize alcohols.

1. 3CH3-CH2-OH + K2Cr2O7 + 4H2SO4 →3CH3-CHO + Cr2(SO4)3 +K2SO4 +7H2O.2. 5CH3-CHOH-CH3 + 2KMO4 + 3H2SO4 →5CH3-CO-CH3 + 2MnSO4 + K2SO4 +8H2O.

Alcohols form esters with both mineral and organic acids:

-+R1 CH2 OH H2SO4 R1 CH2 O SO3

R1 CH2 OH + R COOH R C O CH2 R1

O

Phenols, like alcohols, have hydroxyl groups in theirs structures. The difference is that hydroxyl group in alcohols is bonded to aliphatic structure when hydroxyl group in phenols is bonded to aromatic cycle:

Because of this fact phenols have more acidic properties than alcohols. Phenols react with metals and bases forming phenolates:

OH

NaOH-H2O

ONa

phenol sodium phenolate

Phenols can form esters with organic and inorganic acids. When the acetylsalicylic acid is produced the salicylic acid reacts as phenol and forms ester with the acetic acid:

Phenols are characterized with one specific reaction. They can form coloured compounds reacting with ferric trichloride. In such reaction phenols form coloured coordination compounds - ferric phenolates:

FeCl3 + 6C6H5OH = [Fe(OC6H5)6]3- + 6H+ + 3Cl-.

The reaction with iron(III) chloride solution can be used as a test for phenol. Solutions of various phenols in this reaction acquire their characteristic colour. Phenols will typically yield dramatic purple, blue, red or green colour. This type of reactions takes place when substances containing a phenol group as well as substances containing an enol group react. The enol group is a chemical structure in which there is OH group side-by-side with double bound:

C C

OH

This structure can be found both in aromatic compounds (phenols) and in aliphatic compounds. The reaction with FeCl3 can be used for qualitative determination of all compounds containing OH group attached to the phenol ring as well as for the enol group. Amines are organic compounds containing one or more amino group. There are aliphatic and aromatic amines:

CH3-CH2-NH2

Ethylamine

They are nonpolar and slightly soluble in water substances. Amines show basic properties. Aliphatic amines are more basic than aromatic ones. They can form salts reacting with acids. These salts are polar and soluble in water:

CH3-CH2-NH2 + HCl → CH3-CH2-NH3+Cl-

Experimental

A. Oxidation of ethanol by potassium dichromate or permanganate in the acidic medium

NH2

Aniline

Procedure1. Put 5 drops of ethanol and 1M H2SO4 mixture (2:3) to the first test tube. Add 2 drops of 0.5 M

K2Cr2O7.

2. Put 5 drops of ethanol and 1M H2SO4 mixture (2:3) to the second test tube. Add 2 drops of 0.1 M KMnO4.

3. Put 5 drops of ethanol in the third test tube. Add 10 drops of Fehling-I and 10 drops of Fehling-II to this test tube.4. Shake the contents of the test tubes5. Heat the test tubes over the flame (BE CAREFUL – avoid spilling very strong oxidisers!).

The colour in the first test tube changes from orange to dark-green due to the changed oxidation number of Cr (Cr6+ Cr3+). The violet solution in the second tube becomes colourless and dark MnO2

precipitate forms (Mn7+ Mn4+). If there is an excess of sulphuric acid, the precipitate might dissolve forming colourless MnSO4. The presence of a specific odour (similar to that of apples) indicates that an aldehyde has been formed.The colour in the third test tube does not change and is blue. Explain the differences which are seen in the different test tubes.

B. Reaction of phenol with ferric trichloride FeCl3 In such reaction phenols form coloured coordination compounds - ferric phenolates: FeCl3 + 6C6H5OH = [Fe(OC6H5)6]3- + 3H+ +3Cl-

. Solutions of various phenols in this reaction acquire their characteristic colour. This type of reactions takes place when react substances containing a phenol group as well as substances containing enol group.

ProcedureAdd one drop of 0.1 N FeCl3 into 3 drops of phenol solution. The solution becomes intensely coloured. Record the observed colour.

C. Hydrolysis of acetylsalicylic acid The acetylsalicylic acid is an ester of the salicylic acid, and the -OH group is acetylated. Therefore this compound when treated with FeCl3 does not acquire a colour characteristic for the phenol -OH group. The colour appears when the acetylsalicylic acid is hydrolysed and the phenol -OH group forms instead of the ester group:

Procedure1. Put several crystals of acetylsalicylic acid into a test tube. Add 5-6 drops of water to dissolve them. 2. Add 1 drop of 0.1 M FeCl3. Record the colour.3. Heat the tube a little, the intense violet colour appears. Explain the change of colour.

D. Formation of salts by aniline Aniline displays a basic nature and can form salts in reaction with acids. HCl salt is soluble while the H2SO4 salt has very low solubility. Draw the equations of reactions. Explain these differences.

Procedure.

COOH + H2O

O-CO-CH3

COOH + CH3COOH

OH

Acetylsalicylic acid Salicylic acid

1. Put 1 drop of aniline and 3 drops of water into a test tube, shake it – the emulsion of aniline in water is obtained.2. Immerse a strip of red indicator paper into this emulsion. Record the observed colour.3. Transfer half of the obtained emulsion into another test tube.4. 2 M HCl is added drop-by-drop into the first test tube (1 or more drops) until the transparent aniline hydrochloride is obtained.5. Add 1 drop of 1M H2SO4 into the second test tube and shake it. The precipitation of aniline sulphate should form.

General questions

1. What chemical properties have alcohols?2. What reaction can be used to identify polyhydroxyl alcohols?3. What products form when primary and secondary alcohols are oxidized? What oxidizers should

be used?4. What compounds are formed when alcohols react with acids?5. What are the main differences between alcohols and phenols? Draw the formula of a phenol. 6. What specific reaction is used to identify phenols?7. What other group of substances can react in the same way?8. Why does salicylic acid form coloured compound with FeCl3 and acetylsalicylic acid does not?

Explain this difference.9. Why are amines slightly soluble in water and why are their salts well soluble?

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Conclusion

12. Chemical properties of mono- and polysaccharides

Background. Chemically, carbohydrates are polyhydroxyl aldehydes or polyhydroxyl ketones or substances that yield these compounds when hydrolyzed. Carbohydrates also are known as saccharides. Carbohydrates are classified as monosaccharides, disaccharides, oligosaccharides and polysaccharides according to the number of monosaccharide units linked in a molecule. Monosaccharides usually exist in a cyclic hemiacetal form. The formation of hemiacetal occurs when the oxogroup of monosaccharide reacts with the hydroxyl group of the same monosaccharide. The hemiacetals of monosaccharides can react one with other and form disaccharides and polysaccharides. However, hemiacetals are in equilibrium with the small amounts of open chain aldehyde. Because of this they can be oxidized.

There are formulas of carbohydrates used in laboratory work:

Maltose Amylopectin

Experimental

A. Reaction of hydroxyl groups of carbohydrates Adjacent –OH groups present in saccharides form soluble coloured coordination compound with Cu (II) in alkaline solution:

2 2NaOHCH

CH

OH

OH+ Cu(OH)2 + Cu

O

O

O

O

CH

CH

CH

CH

2Na++4H2O

blue colourprecipitate

water solubleblue complex compound

2-

Procedure:1. Prepare 4 test tubes.2. Put 3 drops of 1% carbohydrate solutions into the test tubes: 1st – glucose, 2nd – fructose, 3rd – sucrose and 4th – lactose or maltose. 3. Add 6 drops of 2M NaOH and 1 drop of 0.1 M CuSO4 solution into each tube. The blue Cu(OH)2 precipitate should form in all the test tubes. After some shaking the precipitate disappears as the blue-coloured coordination compound between the copper and saccharide is formed.Leave the tubes for the use in following experiments!

B. Oxidation of the aldehyde group of carbohydrates. Aldoses exist primarily in cyclic hemiacetal form; however, they are in equilibrium with small amounts of open chain aldehyde. Free aldehyde groups can be easily oxidized to carboxyl by mild oxidizing agents – for example, Cu++ in alkaline solution:

R-CHO + 2 Cu++ + 5 OH- + heating R-COO- + Cu2O + 3H2OBlue solution Red precipitate

Trommer’s reagent: CuSO4 and NaOH;Fehling’s reagent: CuSO4 and basic sodium potassium tartrate.

Procedure 1: Oxidation by Trommer’s reagent1. Add 5 drops of water to the each test tube from the previous experiment.

2. Heat the tubes in the boiling water bath until the reaction mixture changes colour from blue (Cu++) to the yellow (CuOH) or red (Cu2O).3. Explain the change of colour.Procedure 2: Oxidation by Fehling’s reagent1. Take 4 test tubes, put 5 drops of the same carbohydrates as in the A experiment.2. Add 5 drops of Fehling-I and 5 drops of Fehling-II reagents into each tube.3. Heat the test tubes in the boiling water bath until the colours of reduced copper compounds appear.

C. Hydrolysis of polysaccharides Polysaccharides are cleaved gradually: first into oligosaccharides, then – into monosaccharides. This

degradation is catalized by specialized enzymes in vivo, or by acids in vitro. It could be represented by scheme:

Starch (C6H10O5)n Soluble starch (C6H10O5)n’ (where n’<n) Dextrins (C6H10O5)n’’(n’’<n’<n) Blue violet Violet Brown-red

Maltose (C12H22O11) Glucose (C6H12O6)Yellow Yellow

The colours given below the scheme are of compounds which these hydrolysis products form with iodine (I2).

Procedure 1: hydrolysis when catalyst is H + : 1. Take two test tubes. Put 10 drops of 0.5% starch solution into the 1st test tube and add 2 drops of 1M H2SO4.

2. 2nd tube will be the control tube: put 10 drops of 0.5% starch solution and add 2 drops of water instead of acid.3. Heat both test tubes for approx. 7 min in the boiling water bath.4. Dilute the iodine solution in another test tube: add 1 drop of iodine solution into 5 ml of water.5. Take another test tube; put 2 drops of the hydrolyzed starch (1st tube contents) and 1 drop of the diluted iodine solution. Record the observed colour.6. Repeat the same with the control (2nd test tube contents).Compare the both colours. Try to estimate, which stage of the starch hydrolysis was reached in your experiment.

Procedure 2: hydrolysis when catalyst is enzyme:Hydrolysis of starch in human beings begins already in the mouth as saliva contains enzyme called amylase. Thus, your own saliva is a perfect source of this enzyme.

1. Carefully mix 5 drops of starch solution and approximately the same amount of your own saliva in a test-tube.3. Leave this test tube at the room temperature for 3-5 min.4. Analyse it with iodine solution as it was done in the previous experiment.What conclusions can be made about the effectiveness of the biological catalyst as compared to the non-biological one?

General questions

1. What substances are called “carbohydrates”?2. How are carbohydrates classified?3. What chemical structures usually form monosaccharides?4. What chemical reaction should be done to prove the existence of hydroxyl groups in

carbohydrates?5. What oxidizers are usually used to oxidize carbohydrates?6. What substances can catalyze the hydrolysis of polysaccharides?7. What enzyme catalyzes the hydrolysis of polysaccharides?8. What reagents are used to prove the hydrolysis of polysaccharides?

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Conclusion

13. Specific reactions for components of acidic hydrolysis of ribonucleoproteins

Background. Nucleic acids in cells are bound to proteins by non covalent bonds and form conjugated compounds nucleoproteins. The complex is called deoxyribonucleoprotein if protein is joined to deoxyribonucleic acid (DNA), and ribonucleoprotein if protein is joined to ribonucleic acid (RNA). Nucleoproteins are of high molecular mass ranging from 10kDa to several hundred kDa. The complete acidic hydrolysis of nucleoproteins yields a mixture of substances:

Deoxyribonucleoproteins are found in the nuclei of cells and mitochondria. DNA is associated with basic proteins called histones. The interaction is maintained by ionic linkages between anionic phosphate groups of DNA and cationic groups of the side chains of the basic amino acids of histones. Ribonucleoproteins are composed of RNA and various proteins including enzymes, which participate in translation. They are located in the different compartments of the cell including cytoplasm and mitochondria.

Experimental

Yeasts were subjected to hydrolysis for a long time (about 1 h) by boiling with 1M H2SO4. This part of laboratory work was done in advance by the technician. Students are to do the reactions with the prepared yeast hydrolyzate.

Procedure:A. Biuret reaction for peptides

1. Transfer 5 drops of yeast hydrolyzate to the test tube.2. Add 10 drops of 10 NaOH and 1 drop of 1 CuSO4 solution.The violet colour indicates the presence of polypeptides.

B. Specific reaction for ribose – Fehling’s reaction 1. Transfer 10 drops of yeast hydrolyzate to the test tube. 2. Add 10 drops of Fehling’s reagent I 3. Add 10 drops of Fehling’s reagent II4. Heat over the gas flame until a brown-red precipitate appears.

C. Specific reaction for phosphoric acid. 1. Transfer 10 drops of ammonium molybdate solution in HNO3 to the test tube.2. Add 5 drops of yeast hydrolyzate3. Boil the contents of the test tube for a few minutes. The yellow colour appears in the presence

of inorganic phosphate.Reaction: 12(NH4)2MoO4 + H3PO4 + 21HNO3 (NH4)3PO4 12MoO3 + 21NH4NO3 + 12H2O

General questions

1. What compounds form nucleic acids with proteins? What types of them are there?2. What proteins are associated with deoxyribonucleic acids?3. What substances are produced during the complete hydrolysis of nucleoproteins?4. What reaction proves the existence of proteins in the hydrolyzate?5. What reaction proves the existence of carbohydrates in the hydrolyzate?6. What reaction proves the existence of phosphates in the hydrolyzate?

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Conclusion

14. Reactions of amino acids and proteins

Amino acids are heterofunctional compounds:

They have an asymetric C atom, except glycine, so that they have optical isomers. There are L and D series of optical isomers distinguished. Only amino acids of L-series are present in living organism. In amino acids, numeration of atoms begins from carboxylic group that has an adjacent amino group. In carbon atom chain, the second C atom is called -C, and amino group attached to it is called -amino group. In the relation, such amino acids are referred as to -amino acids.Classification:1. According to the shape of C atom chain amino acids fall into aliphatic, heterocyclic and aromatic.2. According to the type of substituents amino acids are grouped into unsubstituted aliphatic,

hydroxylamino acids, S-containing amino acids (mercapto amino acids), diamino acids, amides of amino acids, dicarboxylic amino acids, heterocyclic amino acids, and imino acids. In this classification, the unit:

is considered as the main chain and any group of atoms attached to it is considered as a radical or substituent.

3. According to polarity of molecule and the charge they fall into:a) Nonpolar amino acids including aliphatic, imino acid proline and aromatic amino acids;

Amides

b) Polar amino acids with non-ionized radical, including mercapto and hydrohyamino acids and amides of amino acids;

c) Polar negative charged amino acids, they are dicarboxylic amino acids;d) Polar positively charged amino acids, they are diamino monocarboxylic acids.

It is commonly accepted that names of amino acids are abbreviated using the letter.

Properties of amino acids:

1. Acid-base properties. They depend upon the presence of carboxylic group with acidic properties and amino group with basic properties. In the relation, amino acids are ionized in solutions:

Ionization ability of functional groups is characterized by pKaUnder pH changing, ionization of amino acids is also changed:

Isoelectric point of amino acids (pI) is pH at which the molecule is electrically neutral. The value of pI is determined by titration.2. Condensation reaction. It occurs between amino acids and results in formation of peptide bond:The product is called peptide. Peptide bond is covalent and nonpolar. It is a half double bond located in one plane:

Polypeptides are long chains of amino acid residues linked by peptide bonds. The chain of amino acid residues linked by this bond is the primary structure of proteins. Amino groups of amino acids and peptide bonds of peptides can be identified by specific reactions. You will do them in laboratory works.

Practical

A. Coordination compounds of amino acids with copper salts

Such state of molecule is zwitter ion. The total charge zwitter ion is zero, since positive charge or amino group is compensated by negative charge of carboxylic group.

Amino acids with ions of copper form blue-coloured coordination compounds – chelates. Each molecule of the formed compound consists of two molecules of amino acid and one copper ion:

Procedure:Put one drop of each 0.1 M CuSO4, 4 % glycine and 2 M NaOH solutions into a test tube. The intense blue colour appears which is characteristic for the copper coordination compounds.

B. Biuret reaction for proteins In a very strongly basic medium, copper ions with the peptide bonds form a violet coordination compound. Only peptides containing no less than two peptide bonds and three amino acids can participate in this reaction. It can be used for the identification of proteins.

Tautomeric forms of a peptide bond: C

OHN C

OH

N

ProcedurePut into a test tube 1 mL of the protein solution, 1 mL of 2 M NaOH and 1 drop of 0.1 M CuSO 4. The violet colour with a red or blue tint appears. If there is not much protein then a larger quantity of CuSO4 can be carefully poured along the test tube wall trying not to mix it with the contents of the tube (mixture of protein and NaOH). A violet ring (sometimes called the biuret ring) forms in the test tube.

C. Reaction of proteins with ninhydrine While treated with ninhydrine, -amino acids become oxidized, deaminated and decarboxylated. Ninhydrine formed in the reaction, reduced ninhydrine and NH3 form a blue-coloured compound containing conjugate bond system.

If instead of the amino acid a protein is used in the reaction, then the reaction proceeds without CO2

formation.Procedure

R-CH

O

O

HC-R

C=O

O=C

NH2

NH2

Cu

...-NH-CHR1-C=N-CHR2-CO-N-CHR3-CO-...

...-NH-CHR4-CO-N-CHR5-C=N-CHR6-CO-...

OH

OH

Cu

O

O

OH

OH+

R-CH-COOH

NH2

to

O

O

H

OH+ R-C=O

H+ CO2 + NH3

O

O

OH

OH

O

O

H

OH+ NH3 +

O

O O

O

=N--3H2O

Ninhydrine Reduced ninhydrineBlue-coloured compound

Into 5 drops of the provided protein solution add 5 drops of the 0.1 % ninhydrine solution. Boil the mixture for about 5 minutes. The violet colour appears, and when the solution becomes cold, the colour changes to blue.

This reaction is used for the qualitative determination of proteins and amino acids, in the chromatography of amino acids and colorimetric analysis.

D. Xantoprotein reaction When treated with concentrated HNO3, aromatic amino acids within protein molecule react with the nitric acid, and nitro compounds of yellow colour (xanthos in Greek) are formed. If accidentally some drop of nitric acid contacts with human skin or nails then due to this reaction they become yellow.

Procedure1. Add concentrated HNO3 drop by drop into a test tube with 1mL of undiluted protein solution until the protein precipitate is formed. Also, 5mL of 10 % protein solution and 3 drops of concentrated HNO3 could be used.2. Heat the test tube very carefully. The solution and precipitate gain a yellow colour.

Estimation of molecular mas of a protein by the electrophoresis method

1

+ 2 HNO3

- 2 H2O

Tyrosine

CH2-CH-COOHHO

NH2

O2N

O2N

CH2-CH-COOHHO

NH2

2,4-dinitrotyrosine (yellow)

2

Estimation of molecular mas of a protein from the gel chromatography data

Calculations of molecular mass of a protein by the electrophoresis method and from the gel chromatography data

a, mm Rf, a/h, h= 115mm Mw Log of Mw1 116000 5.0644582 97000 4.9867723 66000 4,8195444 48500 4.6857425 29000 4.462398X

a – distance of the line in the gel from the beginning to the centre of the spoth – distance from the top to the end of gel

Elution volume, ml Mw Log of Mw1 670000 5.8260752 4500000 5.6532133 232000 5.3654884 158000 5.198657X

Elution volume is calculated at the peak of absorption

General questions

1. What substances are called “amino acids”?2. What type of isomers do amino acids form?3. What are the principals of the classification of amino acids?4. What happens to amino acids when pH of solution changes?5. What is the “isoelectric point of amino acids”?6. What bond is called a “peptide bond”? 7. What compounds do amino acids form with copper salts?8. What compounds can participate in the biuret reaction?9. What is the principle of reaction between proteins and ninhydrine? Explain it.10. What amino acids react with the nitric acids?11. What is the principle of the estimation of molecular mass of proteins by the method of

electrophoresis?12. What is the principle of the estimation of molecular mass of proteins using the method of gel

chromatography?

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Conclusion

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PRACTICALS OF MEDICAL CHEMISTRY - Kauno ...julivan/medical-chemistry... · Web view10 April 18 10. Carbonyl compounds – aldehydes, ketones and acids. Their properties – specific