One problem with elimination reactions is that mixtures of products are often formed. For example, treatment of 2-bromo-2-methyl
butane with KOH in ethanol yields a mixture of two alkene products. What are their likely structures?
1 answer:
Answer:
Structures are given below.
Explanation:
- Treatment of 2-bromo-2-methylbutane with KOH in ethanol will give elimination of HBr through E2 mechanism.
- H atoms adjacent to Br will be eliminated.
- 2-bromo-2-methylbutane has two possible adjacent H atoms that can be eliminated giving mixture of products.
- Product of this elimination reaction is alkene. Here saytzeff fule is followed during elimination. So most substituted alkene will be major product.
- Structure of alkenes are given below.
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Answer:
Ka = 6.02x10⁻⁶
Explanation:
The equilibrium that takes place is:
- HA ⇄ H⁺ + A⁻
- Ka = [H⁺][A⁻]/[HA]
We <u>calculate [H⁺] from the pH</u>:
- pH = -log[H⁺]
- [H⁺] =
- [H⁺] = 9.12x10⁻⁴ M
Keep in mind that [H⁺]=[A⁻].
As for [HA], we know the acid is 0.66% dissociated, in other words:
- [HA] * 0.66/100 = [H⁺]
We <u>calculate [HA]</u>:
- [HA] = 0.138 M
Finally we <u>calculate the Ka</u>:
- Ka =
= 6.02x10⁻⁶
Answer:
The amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.
Explanation:
Let's assume amount of sodium is x mg per cracker, y mg per pretzel and z mg per cookie.
So, the following three equations can be written as per given information:
x+y+z = 149 ........(1)
8y+8z = 936 ........(2)
6x+7y = 535 .........(3)
From equation- (2), y+z = = 117
By substituting the value of (y+z) in equation- (1) we get,
x = 149-(y+z) = 149-117 = 32
By substituting the value of x into equation- (3) we get,
y = = 49
By substituting the value of y into equation- (2) we get,
z = (117-49) = 68
So, the amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.