When the phenol shown below is treated with KOH, it forms a product whose IR spectrum does not show an absorption in the 3200-36
00 cm-1 region. Propose a structure for the product.?
1 answer:
Answer:
Explanation:
The first attached diagram shows the proposed phenol that is being treated with KOH to form Oxonine (CₐHₐO).
Reaction of Arylhalide after treatment with KOH lead to the displacement of Bromine atom with the hydrogen from the hydroxyl (OH) substituent , then forming HBr , leaving the oxy (-O) atom alone to form an Oxonine compound.
Therefore the propose structure for the product (Oxonine) CₐHₐO is shown in the second diagram attached below.
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This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.
In this case, it is recommended to write the enthalpy for each substance as follows:
Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and and
are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:
It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:
Finally we convert this result to kJ:
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