In Chapter 4, we will learn that single bonds experience free rotation at room temperature, while double bonds do not. Consider
the two C-N bonds in the structure. One of these bonds exhibits free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. Identify the C-N bond with restricted rotation, and justify your answer by drawing resonance structures.
1 answer:
Answer:
Explanation:
The main objective here is to draw a diagram of an heterocyclic compound containing two C-N bonds in the structure. One with free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. After that ; we will identify the C-N bond with restricted rotation, and also justify our answer by drawing resonance structures.
So; the first image below shows the structure of the heterocyclic compound containing two C-N bonds in the structure with One with free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. From the first diagram. the squared area indicates the C-N bond that exhibits restricted rotation.
The amide bonds in the C-N bonds offers the resonance characteristics and thus exhibits restricted rotation. The resonance is shown in the second image below
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Answer:
a) volume of ammonium iodide required =349 mL
b) the moles of lead iodide formed = 0.0436 mol
Explanation:
The reaction is:
It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.
Let us calculate the moles of lead nitrate taken in the solution.
Moles=molarityX volume (L)
Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol
the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol
The volume of ammonium iodide required will be:
the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol