When the phenol shown below is treated with KOH, it forms a product whose IR spectrum does not show an absorption in the 3200-3600 cm-1 region. Propose a structure for the product.?

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.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :