Write a mechanism for the bromination of acetanilide with bromine in acetic acid. Be as complete as possible and include all rel
evant resonance structures.
1 answer:
Answer:
See figure 1 and 2
Explanation:
In this case, we can start with the general reaction. Acetanilide with bromine in acetic acid we will produce <u>p-bromo-acetanilide</u> (See figure 1).
In the reaction mechanism, we have as first step the attack of the double bond in the benzene ring to one of the Br atoms in bromine () and a positive carbon will be produced in benzene. This positive charge can be moved inside of the benzne ring producing the r<u>esonance structures A, B and D</u>. Resonance structure C is produced when a double bond is generating between the "N" atom and the carbon in the benzene ring.
The final step in the mechanism is when we remove the "H" atom in structure A to produce again the double bond.
See figure 2
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Answer: The percent yield of this combination reaction is 41.3 %
Explanation : Given,
Mass of = 5.0 g
Mass of = 10.0 g
Molar mass of = 23 g/mol
Molar mass of = 71 g/mol
First we have to calculate the moles of and
.
and,
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation will be:
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of
So, 0.217 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 2 mole of react to give 2 mole of
So, 0.217 mole of react to give 0.217 mole of
Now we have to calculate the mass of
Molar mass of = 58.5 g/mole
Now we have to calculate the percent yield of this reaction.
Percent yield =
Actual yield = 5.24 g
Theoretical yield = 12.7 g
Percent yield =
Percent yield = 41.3 %
Therefore, the percent yield of this combination reaction is 41.3 %
Answer:
the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C
Explanation:
Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.
The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀()
Substituting the values into the equation, it becomes
E = σ/ε₀() = 6.3 × 10⁻⁶/8.854 × 10⁻¹²(
) = 7.12 × 10⁵(
) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C
Therefore, the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C