Using phenol, dimethyl sulfate, and naoh, show how you would synthesize methyl phenyl ether.
1 answer:
The synthesis of Methyl Phenyl Ether is shown below,
The synthesis takes place in two steps,
Step 1: Formation of Sodium Phenoxide:
Phenol being more acidic than Alcohols with pKa ≈ 10 can loose its proton attached to oxygen atom, as the resulting phenoxide ion is stabilized by resonance. Therefore, when phenol is treated with NaOH (bBase) it looses proton and forms Sodium Phenoxide.
Step 2: Formation of Methyl Phenyl Ether:
The Phenoxide ion formed in first step when treated with Dimethyl Sulfate produces Methyl Phenyl Ether through a Williamson's Ether synthesis Reaction. Dimethyl sulfate is a well known alkylating agent when treated with phenols, thiols and amines. Dimethyl sulfate readily transfers the methyl group and forms NaSO₄CH₃.
The reaction is as follow,
The synthesis takes place in two steps,
Step 1: Formation of Sodium Phenoxide:
Phenol being more acidic than Alcohols with pKa ≈ 10 can loose its proton attached to oxygen atom, as the resulting phenoxide ion is stabilized by resonance. Therefore, when phenol is treated with NaOH (bBase) it looses proton and forms Sodium Phenoxide.
Step 2: Formation of Methyl Phenyl Ether:
The Phenoxide ion formed in first step when treated with Dimethyl Sulfate produces Methyl Phenyl Ether through a Williamson's Ether synthesis Reaction. Dimethyl sulfate is a well known alkylating agent when treated with phenols, thiols and amines. Dimethyl sulfate readily transfers the methyl group and forms NaSO₄CH₃.
The reaction is as follow,
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Answer:
The answer to your question is 92.7%
Explanation:
Balanced Chemical reaction
3 Zn + Fe₂(SO₄)₃ ⇒ 2Fe + 3ZnSO₄
Molecular weight
Zinc = 65.4 x 3 = 196.2g
Iron (III) = 56 x 2 = 112 g
Proportions
196.2 g of Zinc ------------------ 112 g of Iron
20.4 g of Zinc ----------------- x
x = (20.4 x 112) / 196.2
x = 2284.8/196.2
x = 11.65 g of Iron
% yield =
% yield = 0.927 x 100
% yield = 92.7
Answer:
The number of formula units in 3.81 g of potassium chloride (KCl) is approximately 3.08 × 10²²
Explanation:
The given parameters is as follows;
The mass of potassium chloride produced in the chemical reaction (KCl) = 3.81 g
The required information = The number of formula units of potassium chloride (KCl)
The Molar Mass of KCl = 74.5513 g/mol
Therefore, we have;
1 mole of a substance, contains Avogadro's number (6.022 × 10²³) of formula units
Therefore;
0.051106 moles of KCl contains 0.051106 × 6.022 × 10²³ ≈ 3.077588 × 10²² formula units
From which we have, the number of formula units in 3.81 g of potassium chloride (KCl) ≈ 3.08 × 10²² formula units.