For centuries the Caspian Sea is widely known for its oil fields. Nowadays it attracts ever increasi — Organic Chemistry Chemistry Question
Holy War against Four Horsemen of the Apocalypse
For centuries the Caspian Sea is widely known for its oil fields. Nowadays it attracts ever increasing attention of researches as a treasury of biological resources, in particular, of algae which can be considered as virtually inexhaustible source of unique chemicals. These studies are in line with the global fight of Enlightened Humanity against the four Horsemen of the Apocalypse − Conquest, War, Famine, and Death.
Without revealing the molecular formula, deduce which elements may be found in Y.
Model Answer
Since both X and Y are organic compounds, they contain carbon transformed into carbon dioxide under combustion in an excess of oxygen. CO2 is then reacted with an excess of aqueous calcium hydroxide solution according to the equation:
Ca(OH)2 + CO2 → CaCO3↓ + H2O
Thus, calcium carbonate is the white precipitate (individual compound). This eliminates from consideration a plenty of elements that might potentially be present in Y, including fluorine, chalcogens and phosphorus.
There is no nitrogen in Y, as soon as combustion products are absorbed completely by calcium hydroxide. Lack of condensation of any substance upon cooling of the combustion reaction products from 150 °С down to 0 °С strongly suggests that water is not one of these. In other words, all hydrogen atoms are bound to halogens (molecular hydrogen halides).
It is known that HF and HCl are readily formed under combustion of organic compounds. In the case of HBr there is an equilibrium (4 HBr + O2 = 2 Br2 + 2 H2O), that is strongly shifted towards hydrogen bromide formation at elevated temperatures, whereas HI is readily oxidized providing colored gaseous mixture and products condensation upon cooling.
Thereby, silver bromide is supposed to be the colored precipitate:
CaBr2 + 2 AgNO3 → 2 AgBr↓ + Ca(NO3)2
Further evidence comes from the fact that algae is commonly known as on the splendid natural sources of bromine and iodine. The results of calculation are summarized below:
Y: n(CaCO3) = 1.62 × 10^-2 mol, n(AgBr) = 9.51 × 10^-3 mol, n(C) : n(Br) = 17 : 10
It can be seen that there is no reasonable solution if a colored precipitate is an individual compound (within the constraints of the molar mass and number of atoms given). Consequently, the colored precipitate is a mixture of salts. Also one should bear silver chloride in mind, as it would be colored in the case of co-precipitation with bromide. Furthermore, it is impossible to exclude the presence of oxygen in Y.
So, Y can be composed of C, H, O, Cl and Br atoms.
Without revealing the molecular formula of Y, write down the equation that includes numbers of atoms in the molecule as unknowns. Can this equation be of help in establishing the Y composition?
Model Answer
The expression k = l + m can be written for any fragment Hk(Hal1)l(Hal2)m. Regardless of the even/odd nature of l and m, the equation does not allow us limiting the range of values k, l and m. However, k ≤ 3, and thus l + m ≤ 3. Remembering the molar mass upper limit, one gets three possible combinations with bromine: (1) 1 Br + 1 Cl, (2) 1 Br + 2 Cl, (3) 2 Br + 1 Cl.
Determine the molecular formula of Y using all the data provided.
Model Answer
Thorough inspection of the combinations leads to the "right" ratio of the number of moles of halogen and carbon in the case (1) for Y. For an equimolar mixture of AgBr and AgCl:
187.77 x + 143.32 x = 1.786
where x is the amount of each halogen.
Therefore x = 5.394×10^-3 mol and the molar ratio С : Br : Cl = 3 : 1 : 1.
Thus, Y necessarily contains the C3H2BrCl fragment. Since the mass of the sample is known, one can calculate its molecular weight followed by the number of oxygen atoms in the molecule:
n(O) = [1.000 / (5.394 × 10^-3) - (12.01 × 3 + 1.008 × 2 + 79.90 + 35.45)] / 16.00 = 2.
The molecular formula of Y is C3H2O2BrCl.
Will the substitution of silver nitrate by silver oxide in ammonia solution in the above experiment change the weight and color of the precipitate? To support the answer, calculate the solubility of AgBr (Ks = 5.4 ⋅ 10–13) in NH3 solution (c = 1.0 mol dm-3) taking into account that the two first stepwise formation constants of silver-ammonia complexes are 103.32 and 103.92, respectively.
Model Answer
Yes. Silver bromide and iodide are not completely soluble even in concentrated ammonia, while silver chloride is not precipitated when an ammonia solution of silver oxide is used. A shift of the equilibrium towards formation of complex compounds in the solution undoubtedly changes the mass of the precipitate. Furthermore, due to the difference in the values of solubility products of AgBr and AgCl, the precipitate component ratio will change. This could be behind the color difference (even though not so sharp).
Calculations:
[Ag+][Br-] = 5.4 * 10^-13
[Ag+][NH3] * 10^3.32 = [Ag(NH3)+]
[Ag(NH3)+][NH3] * 10^3.92 = [Ag(NH3)2+]
[Ag(NH3)+] + [NH3] + 2[Ag(NH3)2+] = 1.0
[Ag+] + [Ag(NH3)+] + [Ag(NH3)2+] = [Br-] = s
Strict joint solution of the resulting system of equations (neglecting acid-base equilibria in concentrated ammonia) leads to the value of s = 3.0 ⋅ 10^-3 mol dm^-3.
The same answer can be obtained by making the reasonable assumption that [NH3] ≈ c(NH3) = 1.0 mol dm^-3, because of the expected low solubility of salt. Then the equations would be simplified to the form:
[Ag+] * s = 5.4 * 10^-13
[Ag+] * (1 + 10^3.32 + 10^3.32 * 10^3.92) = s
So, s is still 3.0 ⋅ 10^-3 mol dm^-3, and the assumptions made are valid.
Subjecting 1.000 g of X to the described above analysis sequence results in a colored gaseous (250°С, 1 atm) mixture of products leading to 0.756 g of a white individual precipitate after passing the mixture through an excess of aqueous calcium hydroxide solution. Addition of silver nitrate solution to the supernatant also provides a colored precipitate.
It is known that molecules of Y and X differ by one element, Y having one atom more. X can exist as a mixture of enantiomers, whereas Y reveals geometric isomerism. Furthermore, Y reacts with 0.1 M aqueous KOH solution at room temperature.
Deduce the molecular formula and draw the structure of X.
Model Answer
Y reacts with aqueous KOH solution (c = 0.1 mol dm-3) at room temperature, which is due to the presence of the carboxyl group in its structure. Since both X and Y belong to the same class of organic compounds, X is also a carboxylic acid. Therefore, the compounds differ qualitatively by halogen. The color of combustion products supports the iodine presence in X. The lacking atom in X should have an even valence, which suggests carbon for this position. Since 7.55 · 10-3 mol CaCO3 obtained, M(X) = 1.000 / 0.00755 = 265 g mol-1. Hence its molecular formula is C2H2O2BrI, and the structure is: [VISUAL] (representing CH(Br)(I-COOH, which is confirmed by chirality).
Draw all possible geometric isomers of Y.
Model Answer
Six stereoisomers are possible for Y (a typical case when having four different substituents at a double bond): [VISUAL] (showing six different geometric/positional isomers: (1) E-3-bromo-3-chloroprop-2-enoic acid, (2) Z-3-bromo-3-chloroprop-2-enoic acid, and other isomers arising from alternative positional combinations of H, Cl, and Br on the double bond).
Based on theoretical considerations decide which of the Horsemen of the Apocalypse can be potentially defeated by X and Y?
Model Answer
It is difficult to imagine the use of chemical compounds to prevent large-scale conflicts and wars. Use of X and Y as food resources to beat the famine is unlikely either due to their suspected toxicity for mammals (both are rather strong acids with a strong necrotic effect on mucous membranes). At the same time the potential antibacterial and antiseptic properties of X and Y make them perspective in fighting the horses of Conquest and Death.
Substance Z (41.00 % O by weight) belongs to the same class of compounds as X and Y. However, Z has been only detected in leaves of some plants, and never in algae. Combustion of a 1.000 g sample of Z in a large excess of oxygen followed by complete absorption of colorless gaseous (25°С, 1 atm) products with an excess of aqueous calcium hydroxide solution leads to 3.065 g of a white precipitate. By contrast, no precipitate formed when the supernatant from the previous test was added to an excess of silver oxide in ammonia solution.
It is also known that:
* one of the gaseous products of Z combustion has the density of 1.43 g dm-3 (measured at 34°C, 750 Torr);
* the number of atoms of any element in the molecule of Z does not exceed 3.
Find the molecular and structural formulae of Z.
Model Answer
Assuming that the precipitate is an individual compound (calcium carbonate), let us calculate the molar ratio of carbon and oxygen in the molecule of Z:
n(C) = 3.065 / 100.09 = 3.062 * 10^-2 mol.
n(O) = 1.000 * 0.4100 / 16.00 = 2.563 * 10^-2 mol.
n(C) : n(O) = 3.062*10^-2 : 2.563*10^-2 = 1.195 : 1.000 ≈ 1.2 : 1 ≈ 6 : 5.
However, the number of atoms of each element in Z should not exceed 3, so the initial guess is incorrect.
One can calculate the molar mass of Z gas: M = (rho * R * T) / p = (1.43 g/dm^3 * 8.314 J/(mol*K) * 307 K) / 100.0 kPa = 36.5 g/mol.
This value corresponds to hydrogen chloride, which is further confirmed by the fact that there is no precipitation when supernatant (calcium chloride) is treated with an ammonia solution of silver oxide. However, the presence of chlorine in Z cannot explain the existence of at least two components in the precipitate. Hence, Z must contain fluorine or sulfur atoms (or both elements). Note that HF and SO2 are gases at 25°С and 1 atm, and CaF2 and/or CaSO3 produced in the reactions can precipitate:
Ca(OH)2 + 2 HF → CaF2↓ + 2 H2O (1)
Ca(OH)2 + SO2 → CaSO3↓ + H2O (2)
Since Z belongs to the class of carboxylic acids, it contains 2 or 3 oxygen atoms. Then, the possible values of its molecular mass are:
M = 2 * 16.00 / 0.4100 = 78.0 g/mol or M = 3 * 16.00 / 0.4100 = 117.1 g/mol.
Calculation of the molecular masses of the following combinations of atoms: Mr(CSClO3H) = 128.5, Mr(CSClO2H) = 112.5, Mr(CFClO3H2) = 116.5, and Mr(CFClO2H2) = 100.5 shows that (S + Cl) combination is not suitable because of the upper limit on the molar mass. The (F + Cl) combination is also invalid, since the missing difference of mass cannot be attributed to any of atoms.
Our straightforward solution is deadlocked. The only way to escape consists in assuming that the density of 1.43 g dm-3 may correspond to hydrogen fluoride oligomerized in the gas phase. Then chlorine is excluded from consideration, and only two options remain: Z is composed of only fluorine atoms or both fluorine and sulfur atoms at a time. In the latter case, the molar mass of combination CSFO3H is equal to 112 g mol-1 and that of CSFO2H to 96 g mol-1. The rest of the mass cannot be attributed to any atom.
Then, Z is composed of C, H, F, and O atoms. It contains either two or three oxygen atoms. The molar mass of the compound will be of 78.0 g mol-1 in the former case, and of 117.1 g mol-1 in the latter one. Analysis of possible combinations of atoms provides for molecular formula C2H3FO2 in the former case, and CF3O3 or C4H2FO3 in the latter one. C2H3FO2 is the only correct answer from the chemical point of view. This variant is also supported by calculation of the precipitate mass.
Similarity between Z, X, and Y suggests the presence of a carboxyl group. Then the structural formula of Z can be deduced unambiguously: CH2F-COOH (monofluoroacetic acid).
Both Z and its sodium salt are highly toxic to all mammals.
Knowledge of the area of organisms producing Z as well as of its metabolic pathway in mammals can be considered as a weapon against some of the Horsemen of the Apocalypse. Comment which of these, in your mind.
Model Answer
According to some studies, monofluoroacetic acid and its sodium salt are responsible for the death of approximately 10 % of cattle in South Africa. Death occurs when animals eat the leaves of plants containing monofluoroacetate in high concentration. Thus, the understanding of biological processes associated with monofluoroacetic acid is important to combat Famine currently raging in several areas of the African continent as well as Death.