For many people the taste and odor of truffles are so delightful that truffles cost more than their — Organic Chemistry Chemistry Question
Delightful odor of truffle
For many people the taste and odor of truffles are so delightful that truffles cost more than their weight in gold. The compound X is responsible for the divine smell of the black truffle. The treatment of 0.108 g of compound X with the acidified solution of HgSO4 leads to the formation of some precipitate Z. The treatment of the formed organic compound A with the excess of [Ag(NH3)2]OH afforded 0.432 g of metallic silver. Gas formed as a result of the burning of 0.648 g of compound X was divided into two equal parts. One part was passed through the Ba(OH)2 solution, 3.075 g of precipitate was formed. Another part was passed through NaOH solution. After some time the excess of BaCl2 solution was added. It led to the formation of 3.171 g of precipitate.
Write down the structural formulae of compounds X, Z, A. Determine the weight of the precipitate Z. Assume that all reactions proceed with 100% yield.
Model Answer
The difference between the weights of precipitates formed after direct reaction of gas with Ba(OH)2 and ad a result of stepwise procedure (NaOH, after some time – BaCl2) can be explained only by the presence of SO2 in the gas mixture. Direct reaction with Ba(OH)2 leads to formation of BaCO3 and BaSO3. When gas mixture was passed through NaOH solution, Na2CO3 and Na2SO3 were formed. Sodium sulfite is oxidized with time by oxygen from the air. So, the addition of BaCl2 leads to the precipitation of BaCO3 and BaSO4. Therefore, (3.171 – 3.075) = 0.096 g corresponds to the 6 mmol of additional oxygen atoms, i.e. 6 × 2 = 12 mmol of SO2 was present in the gas mixture. The first precipitate contains 1.773 g (9 mmol) of BaCO3 and 1.302 g (6 mmol) of BaSO3. The second precipitate is formed by 1.773 g of BaCO3 and 1.398 g of BaSO4. Therefore, 0.648 g of compound X contains 0.216 g (18 mmol) of carbon, 0.384 g (12 mmol) of sulfur and 0.648 – 0.216 – 0.384 = 0.048 g (48 mmol) of hydrogen. The brutto-formula of X is C3H8S2. It is also the only possible molecular formula. Formation of 0.432 g (4 mmol) of metallic silver shows that 2 mmol of aldehyde participated in the reaction. It corresponds to 1 mmol of CH2O (compound A). CH2O + 4 Ag(NH3)2OH = (NH4)2CO3 + 4 Ag + 6 NH3 + 2 H2O. As a result, it is possible to conclude that compound X is bis(methylsulfanyl)-methane; thioketal of formaldehyde, CH3SCH2SCH3. It is stable to acid hydrolysis but in the presence of Hg2+ salts it forms Hg(SCH3)2 (compound Z) and formaldehyde. 1 mmol of Hg(SCH3)2 is formed from 1 mmol of compound X. Molecular weight of this compound is 294.6. Therefore, weight of the precipitate Z is 294.6 mg (294 mg and 295 mg are also considered as right answers).