Answer:
See below.
Step-by-step explanation:
Ethers react with HI at high temperature to produce an alky halide and an alcohol.
R-OR' + HI ⟶ R-I + H-OR'
<em>Benzylic ethers</em> react by an Sₙ1 mechanism by forming the stable benzyl cation.
- PhCH₂-OR + HI ⟶ PhCH₂-O⁺(H)R + I⁻ Protonation of the ether
- PhCH₂-O⁺(H)R ⟶ PhCH₂⁺ + HOR Sₙ1 ionization of oxonium ion
- PhCH₂⁺ + I⁻ ⟶ PhCH₂-I Nucleophilic attack by I⁻
If there is excess HI, the alcohol formed in Step 2 is also converted to an alkyl iodide:
ROH +HI ⟶ R-I + H-OH
Thus, benzyl ethyl ether reacts to form benzyl iodide (a) and ethanol (b).
The ethanol reacts with excess HI in an Sₙ2 reaction to form ethyl iodide (c).
Answer:
The graph of this equation is shown in Figure 1. As you can see this is a straight line with negative slope and does not intersect the y-axis. So the ...
Explanation:
The procedure of chromatography .
Answer:
Explanation:
The solution contain 0.01 M concentration of Ba²⁺
0.01M concentration of Ca²⁺
Ksp ( solubility constant) for BaSO₄ = 1.07 × 10⁻¹⁰
Ksp for CaSO₄ = 7.10 × 10⁻⁵
(BaSO₄) = (Ba²⁺) (SO₄²⁻)
1.07 × 10⁻¹⁰ = 0.01 M (SO₄²⁻)
1.07 × 10⁻¹⁰ / 0.01 = ( SO₄²⁻)
1.07 × 10⁻⁸ M = ( SO₄²⁻)
so the minimum of concentration of concentration sulfate needed is 1.07 × 10⁻⁸ M
For CaSO₄
CaSO₄ = ( Ca²⁺) ( SO₄²⁻)
7.10 × 10⁻⁵ = 0.01 (SO₄²⁻)
(SO₄²⁻) = 7.10 × 10⁻⁵ / 0.01 = 7.10 × 10⁻³ M
so BaSO₄ will precipitate first since its cation (0.01 M Ba²⁺) required a less concentration of SO₄²⁻ (1.07 × 10⁻⁸ M ) compared to CaSO₄
b) The minimum concentration of SO₄²⁻ that will trigger the precipitation of the cation ( 0.01 M Ba²⁺) that precipitates first is 1.07 × 10⁻⁸ M