In the presence of sulfuric acid, this alcohol is dehydrated to form an alkene through an E1 mechanism. In the box, draw the maj
or alkene product of this reaction.
1 answer:
Answer:
figure is attached
Explanation:
When we treat alcohol with H₂SO₄ we get elimination as the major product.
As we can see in the given reaction that in step 1 the lone pair of electrons of oxygen attached to the alcohol make a bond with the hydrogen of H₂SO₄.
In the 2nd step H₂O gets detached from the parent ring which generated a positive charge on the ring.
In the 3rd step elimination of hydrogen from the carbon next to the carbonium carbon results into formation of an alkene.
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Answer:
Explanation:
cM 0 0
So dissociation constant will be:
Given: c = 0.15 M
pH = 1.86
= ?
Putting in the values we get:
Also
As
Thus the vale of for the acid is
Answer:
The concentration of H⁺ ions is 0.0165 M.
Explanation:
Let's consider the dissociation of H₂SO₄. In the first step, H₂SO₄ acts as a strong acid, completely dissociating into HSO₄⁻ and H⁺. Therefore, the concentrations of these ions will be the <em>same</em> that the initial concentration of the acid.
H₂SO₄ ⇒ HSO₄⁻ + H⁺
Initial 0.010M 0 0
Final 0 0.010M 0.010M
Now, HSO₄⁻ is a weak acid that will dissociate partially to form H⁺ and SO₄²⁻.
HSO₄⁻ ⇄ H⁺ + SO₄²⁻
To find out the concentration of H⁺ from HSO₄⁻ we will use an ICE Chart. We recognize 3 stages: Initial, Change and Equilibrium, and complete each row with the concentration or change in concentration.
HSO₄⁻ ⇄ H⁺ + SO₄²⁻
I 0.010 0 0
C -x +x +x
E 0.010 -x x x
This quadratic equation has 2 solutions: x₁ = -0.018 and x₂ = 0.00649. Since concentrations cannot be negative, we choose x₂. Then, [H⁺] coming from HSO₄⁻ is 0.00649 M.
The total concentration of H⁺ is:
[H⁺] = 0.010 M + 0.00649 M = 0.0165 M