How do molecules appear in gas

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,

7 0

3 years ago

Do parts a, b and c

Leona [35]

Answer:- (a)The pH of the buffer solution is 3.90.

(b) the pH of the solution after addition of HCl would be 3.60.

(c) the pH of the buffer solution after addition of NaOH is 4.32.

Solution:- (a) It is a buffer solution so the pH could easily be calculated using Handerson equation:

$pH=Pka+log(\frac{base}{acid})$

pKa can be calculated from given Ka value as:

$pKa=-logKa$

$pKa=-log(6.3*10^-^5)$

pKa = 4.20

let's plug in the values in the Handerson equation:

$pH=4.20+log(\frac{0.025}{0.05})$

pH = 4.20 - 0.30

pH = 3.90

The pH of the buffer solution is 3.90.

(b) Let's say the acid is represented by HA and the base is represented by $A^-$ .

Original mili moles of HA from part a = 0.05(100) = 5

original mili moles of $A^-$ from part a = 0.025(100) = 2.5

mili moles of HCl that is $H^+$ added = 0.100(10.0) = 1

This HCl reacts with the base present in the buffer to make HA as:

$A^-+H^+\rightarrow HA$

Total mili moles of HA after addition of HCl = 5+1 = 6

mili moles of base after addition of HCl = 2.5-1 = 1.5

Let's plug in the values in the Handerson equation again. Here, we could use the mili moles to calculate the pH. The answer remains same even if we use the concentrations also as the final volume is same both for acid and base.

$pH=4.20+log(\frac{1.5}{6})$