What is the product of the reaction of hydrobromic acid and 2-bromo-1-butene in the presence of acid and ether?
2 answers:
Answer: The major product of the reaction between Hydrobromic Acid and 2-bromo-1-butene in the presence of ether and acid is 2,2-dibromobutane.
Explanation:
The mechanism of the reaction is supported by the Markovnikov's rule which explains that in the addition reaction of alkenes by hydrogen-halogen compounds, the incoming halogen substituent goes to the more substituted Carbon. It can also be stated that incoming hydrogen atom goes to the Carbon with more Hydrogen atoms.
The only case when the reverse of Markovnikov's rule takes place is when Hydrogen peroxide is present and the addition reagent is HBr.
This case is not like that and it simply follows the Markovnikov's rule.
I'll add an attachment of the reaction to this now.
Answer:
On the reaction the product is 2-2-dibromobutane.
Explanation:
2-Bromo-1 butene is given as in the figure. On the breaking of the double bond 2 local electrophilic and nucleophilic radicals will be formed in the 2-Bromo-1-butene and H-Br respectively.
Due to the Markovnikov Rule the nucleophilic radical of the attacking compound bonds with the carbon atom with least number of H atoms so the product formed will be 2-2-dibromobutane as indicated in the figure.
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Answer : The half-life of the compound is, 145 years.
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = time passed by the sample = 60.0 min
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = 100 - 25 = 75 g
Now put all the given values in above equation, we get
Now we have to calculate the half-life of the compound.
Therefore, the half-life of the compound is, 145 years.
Answer:
[H⁺] = ×
M
Explanation:
pH= - log[H⁺]
=> 3.2 = - log[H⁺]
=> [H⁺] =
∴ [H⁺] = ×
M
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