Draw the mechanism of the slow step that occurs in both first-order substitution and first-order elimination reactions for (R)-3
-bromo-2,3-dimethylpentane in methanol with heat applied. Provide curved arrows in Box 1 to depict the flow of electrons and draw the intermediate in Box 2.
1 answer:
Answer:
see explaination
Explanation:
We are given the (R)-3-bromo-2,3-dimethylpentane and asking to draw the curved arrow which is the showing the mechanism for first-order substitution and first-order elimination reactions. We know the formation of carbocation is the rate determining step in the first-order substitution and first-order elimination reactions.
So in the (R)-3-bromo-2,3-dimethylpentane there is –Br gets removed and formed the tertiary carbocation which is more stable, so the curved arrows in Box 1 to depict the flow of electrons and intermediate in Box 2.
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Answer:
A) 0.065 M is its molarity after a reaction time of 19.0 hour.
B) In 52 hours will react 69% of its initial concentration.
Explanation:
The reaction is first order in :
Initial concentration of =
a) Final concentration of after 19.0 hours=
t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)
Rate constant of the reaction = k =
The integrated law of first order kinetic is given as:
0.065 M is its molarity after a reaction time of 19.0 h.
b)
Initial concentration of =
Final concentration of after t =
Rate constant of the reaction = k =
The integrated law of first order kinetic is given as:
t = 185,902.06 s = ≈ 52 hours
In 52 hours will react 69% of its initial concentration.