School Olympiads in Chemistry. Oxidation of glycerol with chromic acid Benzene and ozone reaction equation
Ozonation of benzene
There are developments of a method for the synthesis of glyoxal by ozonation of benzene with an equivalent amount of ozone with further hydrogenation of the resulting products to obtain glyoxal. Benzene attaches ozone to form triozonide, an extremely explosive substance. Under the influence of water, ozonide decomposes with the formation of three molecules of glyoxal according to the scheme. However, due to the high cost of obtaining ozone and the extreme explosiveness, this method is of no practical value.
Oxidation of glycerol with chromic acid
Another possible method for obtaining glyoxal is the oxidation of glycerol with chromic acid in the presence of sulfuric acid at room temperature. Along with glyoxal, formaldehyde is formed in accordance with the reaction equation:
2Cr2O72-+3NOCH2CH(OH)CH2OH+16H4-4Cr3+3(CHO)2+3H2CO+14H2 (1.6)
The rate of the oxidation reaction increases with an increase in the concentration of hydrogen ions. It is assumed that the active oxidizing form in reaction (1.6) is hexavalent chromium of the singly charged HcrO3- ion. When studying the oxidation reaction of glycerol, free radical ions were found, showing that the oxidation reaction of glycerol with hexavalent chromium can proceed according to the mechanism of both one- and three-electron transfer.
It is assumed that the oxidation of glycerol by hexavalent chromium can proceed according to the following mechanism:
The mechanism involves the formation of an unstable binary complex (1.8), which decomposes at a rate that determines the stage of three-electron transfer to formaldehyde, the free radical ion of glyoxal, and the trivalent chromium ion. The resulting radical ion can undergo further oxidation with hexavalent chromium, giving glyoxal and pentavalent chromium (1.10), or recombine, giving a doubly charged ion (1.10), which is oxidized by a pentavalent chromium ion, giving two molecules of glyoxal and a trivalent chromium ion (1.11). The structure of the binary complex has not been established.
The disadvantage of this method of obtaining glyoxal is the periodicity of the process, the need to clean the resulting mixture from sulfuric acid, chromium compounds and formaldehyde formed during the process.
Reactions of ozone with various aromatic compounds in the temperature range (-40) - (-20)°C according to the reaction rate obey the bimolecular law. The reaction activation energy for benzene is 50 kJ/mol, and the rate of the process increases strongly with increasing medium polarity or in the presence of acidic catalysts.
Let us present data on some kinetic parameters of the reaction of ozone with aromatic hydrocarbons in CCl4 at t = 20°C and the initial concentration of ozone О3 = 10-4¸10-6 mol/l, respectively, the stoichiometric coefficient; rate constant - k, l/mol×s; for: benzene - 3; 6×10-2; naphthalene - 2; 2.4; phenanthrene - 1; 0.8×102; pyrene - 2; 0.8×102; polynaphthalene - 1.6×103; anthracene - 3; 5×103 (first stage) and 43 (second stage). After the addition of the first ozone molecule, the conjugation of benzene and naphthalene is broken, and the following reactions proceed much more easily. A comparison of the rate constants for the reactions of various compounds with ozone shows that aromatic compounds react much more slowly than olefins, and the reaction rate constants increase in the series: benzene< нафталин < фенантрен < пирен < антрацен. Озониды бензола и нафталина - вступают в характерные реакции с HI, NaOH, NH2OH·HCl, подвергаются термическому разложению с образованием пары: альдегид + кислота, а также способны к образованию полимеров.
An assessment of the possible induction effect of previously added ozone on the direction of reactions of the neighboring C=C bond can be considered based on the composition of the decomposition products of naphthalene methoxyhydroperoxides: when heated, the intermediate products are respectively converted into phthalic acid semialdehyde methyl ester and dimethyl phthalate, and the mixture of intermediate products contains up to 80% . Thus, the inductive effect of the ozonide cycle formed in the previous reaction event manifests itself in the preferential formation of a bipolar ion at the carbon atom most distant from the site of attachment of the first ozone molecule.
Ozone reactions without affecting the aromatic nucleus are based on the well-known position that substituents enter into the reaction more easily than the aromatic nucleus in oxidation processes or during attacks by free radicals. For example, the rate constants for substituted benzenes in the CH3 substituent series< CH3-CH2 < (CH3)2 CH - растут симбатно с увеличением числа реакционноспособных атомов водорода в заместителе и уменьшением прочности C-H связи.
Substituted alkylaromatic compounds can react with ozone in two ways: with the formation of hydroperoxides by the chain oxidation mechanism and with the formation of ozonides. Moreover, the first direction is predominant, not the second. The course of the reaction according to the radical mechanism is confirmed by the intense chemiluminescence that occurs when ozone is passed through alkylbenzenes, due to the interaction of peroxide radicals with each other.
Under the action of ozone on anthracene, the main product of the reaction is anthraquinone, the amount of which varies within 20–80%, and the yield of anthraquinone depends on the nature of the solvent, increasing in acetic acid and falling in CCl4. The second product (with a yield of 18÷67%) is phthalic acid - C6H5(COOH)2, and the yield of 4,3-naphthalenedicarboxylic acid - C12H10(COOH)2 is (6÷8)%. Anthracene is known to be readily oxidized by oxygen, forming anthraquinone in high yield. Processes of the same type are observed during the oxidation of polycarbonates and alkylaromatic hydrocarbons with ozone.
Thus, in the reactions of ozone with aromatic hydrocarbons, two types of ozone addition to the C=C bonds of the aromatic nucleus are found: 1) all three oxygen molecules of the ozone are retained and ozonides are formed, which have much in common with olefin ozonides; 2) in the molecule of the new compound, one atom out of three is retained.
The reaction of ozone with aromatic hydrocarbons can be used in the following syntheses:
1) obtaining diphenic acid from phenanthrene:
2) obtaining phthalic dialdehyde and phthalic acid (A. S. 240700 USSR, 1969, BI No. 13), by adding naphthalene the first two molecules of ozone out of five possible, after which the reaction slows down greatly:
3) obtaining glyoxalic acid (AS 235759 USSR, 1969, BI No. 6) based on the lower homologue - benzene according to the reaction:
1.6. Ozone reactions with amines, sulfur and organoelement compounds,
as well as polymers
When ozone reacts with amines, for example, tertiary ones, amine oxides are formed, with a high yield (pat. 437566 England, 1935), as well as nitroxide radicals and other compounds (which are used as modifiers and inhibitors of rubber degradation from O3). Reaction schemes for the interaction of O3 with tertiary, secondary, and primary amines are complex and contain many parallel and sequential reactions. For example, in the reaction of ozone with tributylamine in chloroform, more than 40 intermediate and final reaction products were isolated. The kinetics of the reaction of ozone with amines obeys the bimolecular law and depends on the nature of the solvent.
I. The interaction of O3 with tertiary amines is represented by the following scheme:
1) R3N: + O=O+-O– → R3N+-O-O-O– (O3 is added to the amine to form a product, by analogy with the reaction of O3 with aldehydes saturated with hydrocarbons with multiple bonds);
2) R3N+-O-OO–→ R3N → O + O2; (formation of amine oxides);
3) R2N-(O-O-O-)-C(H2)-RI®R2N=CH-(HO-O-O-)-R®R2N-CHOHRI + O2 (or R2N-CH(-O-O-OH)-RI) (occurs substituent oxidation).
The yield of amine oxides is maximum in solvents in the form of chlorine-containing hydrocarbons and alcohols (CCl4, chloroform, methylene chloride). Also, lowering the reaction temperature (<25 ºС) благоприятно сказывается на выходе оксидов аминов. Использование n-пентана уменьшает выход почти в 10 раз. Например, при озонировании трибутиламина в метаноле образуются (в %): (C4H9)3N → 0÷53; C4H9N=CH-C3H7 → 2; C4H9NCH=0 → 3; C4H9NCH=CHC2H5 → 11; (C4H9)2NH → 9; C4H9NCOC3H7 → 6.
II. The reaction of O3 with secondary amines leads to the formation of nitroxy radicals, which, depending on the structure of the amine, can be the main reaction products or be present in appreciable amounts. Aromatic amines and p-phenylenediamine derivatives form nitroxide radicals especially easily. For example, the interaction of ozone with triacetonamine, resulting in a nitroxyl radical (2,2,6,6-tetromethyl-4-oxopiperidoxyl), is highly stable and remains for months at room temperature without noticeable changes. Most aromatic amines are antiozonants and are used to protect rubber products from ozone aging.
The reaction of ozone with secondary amines can be represented according to the scheme (the action of O3 on di-tert-butylamine in pentane, at t = -120 ºС):
III. The main products of the interaction of ozone with primary amines are nitro compounds and ammonium bases. Their relative content depends mainly on the nature of the solvent. When passing from hydrocarbons to chlorine-containing solvents, the yield of nitro compounds decreases, but the yield of ammonium salts increases, i.e., the solvent molecule is involved in the reaction.
The scheme of interaction of O3 with a primary amine can be represented in a general form by the equation:
C4H9NH2 + O3 → C4H9NO2 + O2.
The formation of the final nitro compound requires the consumption of 3 ozone molecules. For comparison, the rate constants of the reaction of ozone with ammonia in aqueous solutions (k = 39 l/mol) are noticeably lower than those of amines (for example, for aniline, k = 2.5 103 at t = 20 ºC).
The main stages of the reaction of tributylthiourea and its analogues with ozone can be represented by a simplified scheme:
Nitroxyl radicals react most readily. Absorbing 1 mole of ozone, they are converted mainly into nitro compounds.
In the reaction of ozone with sulfur compounds, for example, sulfides (R-(–S–)n-R), thioureas and thiosemicarbazides (R-(R)-C=S), the reactions proceed mainly at the sulfur atom. To carry out the reaction with disulfides and polysulfides, a solution in carbon tetrachloride is used. In this case, the initial sulfides quite easily react with ozone with a rate constant k = 103 L/mol s, which is close to that of phenols and much higher than the rate of oxidation of the -CH2- group in alkyl substituents. The main product of the first stage of the reaction is sulfoxide (=S=O), which can then be oxidized to sulfone (=S(=O)2), but at a much slower rate (50–100 times). The rate constants during the interaction of ozone with sulfides, using the example of dimethyl sulfide (CH3-S-CH3) - 1.5 103 l / mol s, compared with sulfur (S8) - 5.5 and ethyl alcohol (CH3CH2OH) - 10. Moreover, there is a decrease in the reactivity of organic sulfur compounds in the series: R-S-R, R-(S)2-RS8.
Ozone also interacts with organoelement compounds, for example, silicon:
(C2H5)3Si-CH2-CH3+O3 ® (C2H5)3Si-CH-(OO )-CH3 + OH ®(C2H5)3SiOOH + O=CH-CH3
or according to the second reaction: ® (C2H5) 3Si-(-O-O-O) -CH2 ® (C2H5) 3SiO2 + OOCH2CH3.
Under the action of ozone on polymeric materials, a particularly strong effect occurs on elastomers containing C=C bonds in the main chain of the macromolecule (for example, rubbers). Under the action of O3 on polymers having a saturated hydrocarbon chain, especially on their solutions (in CCl4 at t = 20 ºC), a drop in molecular weight and accumulation of oxygen-containing functional groups (acids, ketones and peroxides) are observed. Polymers containing phenyl rings in the main chain react most slowly with ozone, while polycyclic (polynaphthylenes, polyatracenes) or polymers with heteroatoms (polycarbonate) react much more easily. In a series of polymers with a saturated hydrocarbon chain, the reaction rate increases from polyisobutylene to polyvinylcyclohexane, while a decrease in the number of chain breaks is observed. Polybutadiene and polyisoprene have the largest rate constant, and they also have the smallest number of discontinuities per reaction act. Some polymers are insoluble in common solvents (eg polyethylene). Ozonation differs from the scheme of thermal-oxidative destruction of polystyrene in that low temperatures and high rates of radical formation create conditions in which the share of chain processes is 15–20% in the reaction balance, and the main part of the products is formed during the decomposition of peroxy radicals. Acids make up a small part of the reaction products and can be formed both as a result of the oxidation of phenoxy radicals or their transformation products, and as a result of the destruction of aromatic ozonides. The action of ozone on other polymers (polyethylene, polyvinylcyclohexane) is accompanied by the formation of peroxide radicals. The destruction of unsaturated polymers under the action of O3 (for example, rubbers, rubbers) occurs similarly to monomers, i.e., along C=C bonds.
4(CNO) 2 + 2HNО 3 → 4СНОСООН +N 2 O+ H 2 O. (1.5)
To initiate the start of the reaction, the presence of sodium nitrate or nitrite is necessary, the interaction of which with nitric acid releases the amount of nitrous oxide necessary to start the process. Thus, sodium nitrate or nitrite plays the role of a catalyst. During the oxidation process, the addition of sodium nitrate (or another compound) is not required, since a sufficient amount of nitrous oxide is formed as a result of the interaction of acetaldehyde with nitric acid.
Selenium dioxide SeO 2 is used as a highly selective catalyst for the conversion of acetaldehyde to glyoxal. In the presence of a selenium-containing catalyst, an increase in the conversion of acetaldehyde from 4 to 10% is observed. However, the regeneration of SeO 2 is a rather serious problem, due to the fact that in the course of the oxidation process, difficult-to-reduce organic selenium compounds are formed. To increase the yield of the target product, formic and acetic acids are added to the reaction mixture in a molar ratio with nitric acid from 0.5 to 1.16. This makes it possible to increase the yield of glyoxal up to 50%. It should be noted that for this reaction, the optimal temperature range lies within a narrow range of 45 ÷ 48 o C. A further increase in temperature leads to a noticeable decrease in the yield of the target product.
Salts of metals of Group I of the Periodic Table are proposed as catalysts: sodium, lithium, silver nitrates. The resulting mixture of products is an aqueous solution of aldehydes, acetic acid, nitric acid and a small amount of nitrous acid. Aldehydic compounds include glyoxal, acetaldehyde, and glyoxalic acid.
The disadvantage of this method of obtaining is the periodicity of the process. In addition, the most important problem is the purification of the resulting mixture of various products. To obtain a commercial product (40% aqueous solution of glyoxal), it is necessary to remove nitric and acetic acids from the mixture, for example, by ion exchange. The isolation of glyoxal from the reaction products by standard methods is impossible. In world practice, this technology for producing glyoxal is used very limitedly due to large emissions of nitrogen oxides into the atmosphere, which destroy the ozone layer that protects the earth's surface from UV radiation.
1.2.2 Ozonation of benzene
There are developments of a method for the synthesis of glyoxal by ozonation of benzene with an equivalent amount of ozone with further hydrogenation of the resulting products to obtain glyoxal. Benzene attaches ozone, forming triozonide - an extremely explosive substance. Under the action of water, ozonide decomposes with the formation of three molecules of glyoxal according to the scheme:
However, due to the high cost of obtaining ozone and the extreme explosiveness of this method, this method is of no practical value.
1.2.3 Oxidation of glycerol with chromic acid
Another possible method for obtaining glyoxal is the oxidation of glycerol with chromic acid in the presence of sulfuric acid at room temperature. Along with glyoxal, formaldehyde is formed in accordance with the reaction equation:
2Cr 2 O 7 2- + ZNOSN 2 CH (OH) CH 2 OH + 16H 4 ↔ 4Cr 3- + 3(CHO) 2 + ZN 2 CO + 14H 2 O(1.6)
The rate of the oxidation reaction increases with an increase in the concentration of hydrogen ions. It is assumed that the active oxidizing form in reaction (1.6) is hexavalent chromium of the singly charged ion HcrO 3 - . When studying the oxidation reaction of glycerol, free radical ions were found, showing that the oxidation reaction of glycerol with hexavalent chromium can proceed according to the mechanism of both one- and three-electron transfer.
It is assumed that the oxidation of glycerol by hexavalent chromium can proceed according to the following mechanism.
Among the various reactions that aromatic compounds enter into with the participation of the benzene ring, the substitution reactions discussed above attract attention first of all. This happens because they proceed contrary to expectations. With the degree of unsaturation that is inherent, for example, in benzene, addition reactions should have been more characteristic of this hydrocarbon. Under certain conditions, this happens, benzene and other arenes add hydrogen atoms, halogens, ozone and other reagents that can add.
11.5.5. Hydrogenation. In the presence of hydrogenation catalysts (platinum, palladium, nickel), benzene and its homologues add hydrogen and turn into the corresponding cyclohexanes. So, benzene is hydrogenated over a nickel catalyst at 100-200 0 C and 105 atm .:
Hydrogenation of arenes has two peculiarities compared to alkenes. Firstly, arenes are significantly inferior to alkenes in reactivity. For comparison with the conditions for the hydrogenation of benzene, we point out that cyclohexene is hydrogenated into cyclohexane already at 25 0 C and a pressure of 1.4 atm. Secondly, benzene either does not add, or attaches three hydrogen molecules at once. It is not possible to obtain partial hydrogenation products, such as cyclohexene or cyclohexadiene, by hydrogenation of benzene.
These features during hydrogenation, a special case of addition reactions to the benzene ring, are due to the structure of benzene. Upon conversion to cyclohexane, benzene ceases to be an aromatic system. Cyclohexane contains 150.73 kJ more energy (resonance energy) and is less stable than benzene. It is clear that benzene is not inclined to pass into this thermodynamically less stable state. This explains the lower reactivity of benzene with respect to hydrogen compared to alkenes. Accession to the aromatic system is possible only with the participation R-electrons of a single electron cloud of the benzene ring. When the addition process begins, the system ceases to be aromatic and a particle rich in energy and highly reactive is obtained, which is much more likely to enter into the addition reaction than the original arene.
11.5.6. Halogenation. The result of the interaction of halogen with benzene depends on the experimental conditions. Catalytic halogenation leads to the formation of substitution products. It turned out that ultraviolet light initiates the addition of halogen atoms to the benzene nucleus of arenes. Benzene itself in the light attaches 6 chlorine atoms and turns into hesachlorocyclohexane, which is a mixture of 9 spatial isomers
One of these isomers, in which 3 chlorine is occupied by axial bonds, and another 3 - by equatorial bonds (γ-isomer, hexachlorane), turned out to be an effective insecticide, a means of controlling harmful insects. Hexachlorane proved to be too stable in the biosphere and capable of accumulating in the adipose tissue of warm-blooded animals, and therefore is not currently used.
In terms of its reactivity with respect to halogens in addition reactions, benzene is significantly inferior to alkenes. For example, chlorine and bromine in carbon tetrachloride, even in the dark at room temperature, add to cyclohexene. Under these conditions, benzene does not react. This happens only under ultraviolet light.
11.5.7. Ozonation. Ozonation is another example showing that benzene, as an unsaturated compound, can enter into an addition reaction. The ozonation of benzene and the study of triozonide hydrolysis products were carried out as early as 1904 ( Harries)
Interesting results were obtained with ozonation about-xylene (1941, Vibo). The fact is that the composition of ozonation products depends on the position of double bonds in the benzene ring. Structure 1 with double bonds between the carbons of the benzene ring bearing methyl substituents, upon ozonation and hydrolysis of the ozonide, will give 2 molecules of methylglyoxal and a molecule of glyoxal
Alternative structure II for about-xylene would have to form 2 glyoxal molecules and a diacetyl molecule
All-Russian Olympiad for Schoolchildren in Chemistry - 2004
"OPTIONAL TASKS"
PHYSICAL CHEMISTRY
Task 1.
"How innocently they would say the year before -
yes, the century before last."
(Bakhyt Kenzheev)
“In the beginning of 1880, Victor Meyer began very interesting experiments on the vapor densities of halides at very high temperatures. The experiments were then repeated by Crafts. different for different halides.
| Temperature | Density chlorine vapor | Density bromine vapor | Density pair of iodine |
| Below 440 about | 2,45 | 5,52 | 8,78 |
| 440 about | norms. | norms. | 8,72 |
| 900 about | norms. | norms. | 8,11 |
| 1200 about | norms. | 4,5 | 6,07 |
| 1400 about - 1500 about | 2,02 | 3,5 | 5,31 |
Assuming that the observed decrease in halide vapor densities does not depend on a significant change in the expansion coefficients of gases at these temperatures, the data presented can be explained ... "
("Essay on the development of chemical views" by N. Menshutkin, St. Petersburg, 1888. pp. 301-302)
1. How can such a significant change in the "vapour density" for halogens be explained?
2. Estimate the error in determining the "vapor density" (relative %).
3. Calculate the composition of the "steam" for iodine at the indicated temperatures (mole fractions).
4. Calculate the binding energy I - I (kJ/mol).
5. Determine the melting and boiling point of iodine (o C) and calculate the lattice energy of iodine if the vapor pressure over iodine is 1 mm Hg (133.32 Pa) at 43.7 o C; 10 mmHg Art. at 77.0 about; 100 mmHg at 122.4 about; 400 mmHg at 162.8 o.
Task 2.
HX- one of the strongest organic acids can be obtained according to the scheme:
Information on the composition of the connections shown in the diagram:
Substance |
|||||
Recently, interest has increased in the use of molecular fluorine in solvents. In this case, it is possible to control the oxidative activity of F 2 by introducing various substances into the solution. Acid HX applies to such compounds. Alleged processes with her participation:
The Nernst equation for molecular fluorine has the form:
Solvents used (solv): CH 3 CO 2 H, HCO 2 H, CF 3 CH 2 OH, CH 3 OH, CF 3 CO 2 H.
1. Identify the unknown substances in the diagram and write the reaction equations.
2. Assuming that the total concentrations of fluorine (C 1) and acid (C 2) are constant (C 1<0,5С 2):
a) Express E 0 in terms of E 0 (F 2 /F -) and K a (HF). (We put E 0 =E 0 (F 2 ,H + /HF), E=E(F 2 ,H + /HF).)
b) Express E as a function of C 2 and C 1 if pH(C 2 , C 1 , solv) is known.
(Admissible approximations make it possible to dispense with K 1 -K 3 .)
3. To the solution (p2) was added (C 3): a) BF 3 (C 3<0,5C 1);
б) NaX(C 3 <
5. A suitable solvent maintains a constant potential in the process of passing fluorine into the solution and forms with F 2 only compounds that are easily separated from the target product. Suggest a suitable solvent from the list, justify.
6.Suggest one way to obtain F 2 in the laboratory without resorting to electrolysis.
7. Why is it impossible to accurately determine the potential of fluorine during the experiment?
Task 3.
Ozonation of benzene
The ozonation reaction is used in organic chemistry for the synthesis of various classes of compounds and the determination of the structure of unsaturated compounds.
Ozonation of benzene proceeds in methyl chloride at –80 o C. This reaction can be described by the kinetic scheme:
1) Write the structural formulas of ozonides А 1 , А 2 , А 3 . What substance is formed when A 3 is reduced with zinc?
2) Another way to obtain ozonides is the dehydration of dihydroxy peroxides of the type
phosphoric anhydride. Write schemes for obtaining mono-, di- and polyozonide from the indicated dihydroperoxide.
One of the qualitative reactions to peroxide compounds of the most diverse structure is the interaction of their ether solutions with a solution of titanyl sulfate in 60% sulfuric acid.
3) What is the analytical signal and what causes it?
Consider the above kinetic scheme. Let us assume that ozone is taken in a slight excess compared to the overall equation.
4) a) On one graph, depict the curves of the dependence of the concentrations of substances A 1, A 2, A 3 on time, assuming that k 1
k 2 k 3 .
b) On one graph, depict the dependence of the concentration of substance A 1 on time in two cases: 1) k 1 << k 2 ;
2) k 1 k 2. c) What do you think, which of the two approximations - (b1) or (b2) - is more true? Why?
d) Express the rate of formation of A 3 in terms of the concentrations of ozone and benzene, provided that the concentrations of intermediates A 1 and A 2 are stationary. What is the overall order of the reaction?
The ozone concentration in the solution can be maintained constant by continuously passing an ozone-oxygen mixture through the solution. Let us consider the kinetics of the reaction under these conditions.
5) a) Express the rate of formation of A 3 in terms of the concentrations of ozone and benzene, provided that the concentrations of A 1 and A 2 are stationary. What is the overall order of the reaction?
b) How many times will the half-life of benzene change with an increase in its initial concentration by 2 times?
c) Solve the kinetic equation from (5a) and find the dependence of the concentration of the product A 3 on time. Designate the initial concentrations of benzene and ozone as 0 and 0.
Task 4.
Benzoyl peroxide (PB) and 2,2'-dimethyl-2,2'-azodipropanoic acid dinitrile (2,2'-azo-bis-isobutyronitrile, AIBN) are standard initiators of radical-chain processes that readily decompose homolytically even at a slight increase in The activation energy of decomposition in inert solvents is 129 kJ/mol for PB and 130 kJ/mol for AIBN, and the pre-exponential in the Arrhenius equation ( k = A e- E a/ RT) BUT= 10 14.5 s –1 for PB and 10 15.0 s –1 for AIBN. Decay proceeds in the 1st order.
Questions and tasks.
1. Write down the formulas for PB and AIBN and the equations for the reactions of their decomposition in an inert solvent. What products can be formed in this case? Name them.
2. C–N and N=N bonds in azo compounds are quite strong (295 and 420 kJ/mol, respectively). Why, then, does AIBN easily form radicals when heated?
3. In one of the experiments, during the decomposition of a blue solution of AIBN, 0.5 ml of gas was released within 1 minute (measured at room temperature and a pressure of 735 mm), and after the end of the reaction, 1.250 liters of this gas were released, measured under the same conditions. Calculate the AIBN decay rate constant under experimental conditions (indicate the time for it in seconds).
4. Estimate the time for which AIBN in solution will decompose by 0.1% at a temperature of 25 ° C, as well as the half-life. At what temperature will AIBN decay by 50% in 5 hours?
5. Estimate the thermal effect of AIBN decay. Is its explosive decomposition possible, and if so, under what circumstances? The breaking energy of a triple bond in a nitrogen molecule is 945 kJ/mol; the C–C bond energy is assumed to be 340 kJ/mol.
6. In one of the works, the initial rate of PB decomposition in boiling benzene was measured. If, according to the data obtained, the decomposition rate constant is calculated under the assumption of the 1st order of the reaction, then it turns out that the constant calculated in this way depends on the initial concentration of PB:
To explain these results, it was assumed that a bimolecular reaction of PB decomposition could proceed in parallel. Calculate the true rate constant of the monomolecular decomposition of PB from the experimental data (it is recommended to use a graphical method).
7. When the decomposition of AIBN was carried out in xylene at 108 ° C in the presence of 2,6-dimethyl- P-benzoquinone, the spectrum of electron paramagnetic resonance (EPR) of the resulting solution showed the presence of stable free radicals; 7 equally spaced lines in the spectrum with a splitting of 0.573 mT (millitesla) indicated the presence of 6 equivalent protons in the radical, and the splitting of each line into three (with a distance between them of 0.137 mT) indicated the presence of two equivalent protons (T.L.Simandi et al., European Polymer Journal, 1989, volume 25, pp.501–507). Using these data, depict the structure of the resulting radical and confirm it with EPR data (the splitting in the spectrum is proportional to the density of the unpaired electron on a given atom). Explain the stability of the resulting radicals. Why do you think the authors published the article in the indicated journal?
Instruction. The dependence of concentration on time for the reaction of the 1st order: With
= With o e –kt or ln( c o / c)
= kt. Gas constant R= 8.31 J/(mol K).
CHEMISTRY AND LIFE
Task 1.
Lake Galichskoe (Kostroma oblast, see map) has a fairly silty bottom, 25 km long, 5 km wide, average depth 1.5 m, water exchange rate ~ 1 time per year. Several rivers flow into the lake, of which the most abundant (if you can talk about those miserable streams) are Srednyaya and Cholsma. Falls out of the lake. Veksa. The town of Galich stands on the lake (20 thousand inhabitants, a truck-crane plant, a tannery, a bakery, a shoe and clothing factory).
Water samples were taken at four points (see map). Samples were taken from 3 to 8 August 2003. The results of the analyzes are shown in the table.
| Dot | ||||
| Temperature | ||||
| pH | ||||
| Transparency, cm | ||||
| Chromaticity, hail | ||||
| Alkalinity total, mM | ||||
| Alkalinity free, mM | ||||
| Hardness (Ca 2+ + Mg 2+), mm | ||||
| Calcium, mM | ||||
| Chlorides, mM | ||||
| About 2,% of saturation |
Note: free alkalinity - the concentration of bases that give pH> 8.2, total alkalinity - the concentration of all bases that can be titrated with hydrochloric acid. The difference between total and free alkalinity is usually due to hydrocarbons.
It can be seen from the results that, firstly, the pH of the water in the lake is abnormally high, and secondly, the concentration of salts in the lake is about 3 times less than in the rivers that feed it. Two explanations have been offered for both facts. First, the residents of the city actively wash clothes, which leads to sodium carbonate and phosphate getting into the lake, alkalizing the water and binding calcium to magnesium. Second, the tannery dumps effluent containing lime into the lake.
1. Give the reaction equations that lead to a decrease in the concentrations of cations and anions in the lake compared to the rivers feeding the lake according to the first and second hypotheses. Write the equations in ionic form.
2. Is the decrease in the concentration of salts in Lake Galich compared to the rivers that feed it local (only within the city) or throughout the lake? Justify your answer with one sentence.
3. What facts do not fit into the first hypothesis of pH increase?
4. What facts do not fit into the second hypothesis of pH increase?
5. What facts do not fit into the second hypothesis of salt concentration decrease?
There is also a hypothesis that the decrease in the concentration of salts in the lake compared to the rivers that feed it is due to natural causes.
6.What could be the reason? Write the corresponding reaction equation.
7. How can sludge analysis confirm or disprove this hypothesis?
8. Estimate the thickness of the layer of these substances deposited over the year, if we assume that only they are deposited, and their density is approximately equal to 2000 g / dm 3?
For reference: K a (H 2 O + CO 2) = 4.5? 10 -7, K a (HCO 3 -) = 4.8? 10 -11, K a (H 2 PO 4 -) \u003d 6.2? 10 -8, K a (HPO 4 2-) = 5.0? 10 -13, PR (CaCO 3) = 4? 10 -9, PR (MgCO 3) = 2? 10 -5, PR (Ca 3 (PO 4) 2) = 2? 10-29 .
Task 2.
Insects are too small to look for a mate in the world around them with the help of their eyes, as a person usually does. For this purpose, they secrete special substances - sexual attractants, or pheromones. A few milligrams of pheromone secreted by a female of some butterfly is enough to attract hundreds of males from a distance of tens of kilometers. At the same time, minor changes in the structure of the pheromone (the position or stereoisomerism of the double C=C bond, the use of another stereoisomer of a chiral molecule, etc.) can lead to the attraction of insects of a completely different species or to the repelling of individuals of their own species.
It is known that the female olive fly Bacroceraoleae attract males with a compound BUT, and males of females - using the connection B. When these compounds are treated with a dilute acid solution, the same substance is formed AT, about which the following is known:
a) it contains 27.6% oxygen and 62.1% carbon, and upon treatment with phosphorus pentabromide, it turns into compound D containing 71.75% bromine;
b) when heated in the presence of catalytic quantities pair-toluenesulfonic acid (TsOH) is easily converted into a mixture of compounds BUT and B;
c) as a result of the chain of transformations:
it can be transformed into a connection W, which, when reacted with one equivalent of a Grignard reagent, forms a mixture of two alcohols.
Define possible connection structures A - Z and write the equations for the reactions given here
Task 3.
Immobilized enzymes
Enzymes are biological catalysts of protein nature. For the use of enzymes in various biotechnological processes, it is often necessary to fix (immobilize) the enzyme on a suitable insoluble carrier (the original enzyme that is subjected to immobilization will be called the native enzyme).
A. Obtaining immobilized enzymes.
Enzymes are usually immobilized using carriers containing amino, hydroxyl, or carboxyl groups, which are not highly reactive under physiological conditions. Therefore, when a covalent bond is formed, the carrier enzyme must first be activated. One of the types of activating reagents are dialdehydes, for example, succinic
OHC-(CH 2) 2 -CHO.
1. Write the reactions that take place under the action of succinic aldehyde on polyvinyl alcohol in an acidic medium. Use the following icon to designate the media here and below:
2. Write an equation for a side reaction between polyvinyl alcohol and succinic aldehyde that reduces the yield of the enzyme covalently bound to the carrier.
3. Write the interaction reaction of the carrier activated with succinic aldehyde with the enzyme (pH = 8). In the diagram, indicate the functional group of the enzyme that will interact. Use the icon to designate an enzyme:
4. The side groups of what amino acid residues will carry out the reaction? Give the trivial names and formulas of the side radicals of these amino acids.
5. What other enzyme group, besides those indicated in the answer to question 4, can interact with the activated carrier?
6. Indicate the pH ranges in which the enzyme-carrier bond formed:
a) stable
b) unstable
To increase the stability of the carrier enzyme bond, the system resulting from the reaction described in question 1 is treated with sodium borohydride.
7. Give the scheme of this reaction. where V max \u003d k 2 [E] 0 and K M \u003d (k 2 + k -1) / k 1
Immobilization can affect the values of both the catalytic constant k2 and the Michaelis constant KM, which characterizes the efficiency of enzyme binding to a given substrate. Catalysis by immobilized enzymes can proceed in two modes:
kinetic, in which the observed rate is determined by the catalytic properties of the enzyme itself;
diffusion, in which the observed reaction rate is controlled by diffusion (that is, it is determined by the rate of substrate supply to the enzyme).
Let us consider a system with an immobilized enzyme, in which a regime change is observed at a substrate concentration equal to [S] 0,ex.
8. Draw on the graph in the so-called double inverse coordinates (1/v from 1/[S] 0) the dependence for this system in the range of substrate concentrations from [S] 0.ex /2 to 10[S] 0.ex. Mark on the x-axis the points corresponding to the substrate concentrations [S] 0.ex /2, [S] 0.ex and 10[S] 0.ex. The dependence corresponding to the kinetic mode, mark the number "1", and the diffusion mode - the number "2".
9. Indicate the length of the segments cut off on the abscissa and ordinate axes in double reciprocal coordinates by continuing the dependence for the case of the kinetic regime (graph from the answer to question 8). Give calculations.
Immobilization on polyelectrolyte supports can affect the distribution of protons in the system, which leads to a change in the dependence of the catalytic activity of the enzyme (and, hence, the reaction rate) on pH (hereinafter, such a dependence will be called the pH profile). Let the pH profile for some native enzyme have a classical bell-shaped form (see below).
10. Present in coordinates (v from pH) the pH profiles corresponding to the native enzyme (label each branch as 1), the same enzyme immobilized on a polyanionic carrier (label each branch as a 2), and the same enzyme immobilized on a polycationic carrier (mark each branch with a 3). Assume that immobilization does not affect:
a) on the value of the rate of the enzymatic reaction in the pH optimum (the highest point of the bell);
b) on the shape of the pH profile.
