Using your knowledge of reactions and mechanisms, draw the mechanism for the formation of 2-methylpropene from 2-methyl-2-propan
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1 answer:
Answer:
Here's what I get
Explanation:
The reaction is an E1 elimination of an alcohol to form an alkene. It has three steps:
1. Protonation
The alcohol is protonated with aqueous sulfuric acid to convert it into a better leaving group.
2. Loss of the leaving group
A water molecule leaves in a unimolecular process to form a stable 3° carbocation.
3. Loss of an α-hydrogen
A water molecule removes an α-hydrogen, forming 2-methylpropene and regenerating the original hydronium ion.
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pH is the hydrogen ion concentration and pOH is the hydroxide ion concentration in the solution. pH KOH is 12.73, pOH KOH is 2.24 and pH NaCl is 7.
<h3>What are pH and pOH?</h3>pH is the negative log of the hydrogen ion concentration and pOH is the negative log of the hydroxide ion concentration.
The relation between the pH and pOH can be given as,
The pH of KOH can be calculated by the formula,
In the first case, the concentration of the KOH is
Substituting values in the equation:
Hence, the pH of KOH is 12.73.
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pOH of KOH can be calculated by the formula,
The hydroxide concentration of the KOH solution is
Substituting value in the equation:
Hence, the pOH of KOH is 2.24
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The pH of NaCl can be calculated by the formula,
In the third case, the concentration of the NaCl is
Substituting values in the equation:
Hence, the pH of KOH is 7.0.
Therefore, KOH is basic and NaCl is approximately neutral.
Learn more about pH and pOH here: