Compounds X and Y are both C6H13Cl compounds formed in the radical chlorination of 3-methylpentane. Both X and Y undergo base-pr
omoted E2 elimination to give a mixture of alkenes.In water X and Y each react to form a mixture of substitution and elimination products; however, X reacts much faster than does Y.What is the structure of Y?
1 answer:
Answer:
Y is a 3-chloro-3-methylpentane.
The structure is shown in the figure attached.
Explanation:
The radical chlorination of 3-methylpentane can lead to a tertiary substituted carbon (Y) and to a secondary one (X).
The E2 elimination mechanism, as shown in the figure, will happen with a simulyaneous attack from the base and elimination of the chlorine. This means that primary and secondary substracts undergo the E2 mechanism faster than tertiary substracts.
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Answer:
5.41 g
Explanation:
Considering:
Or,
Given :
For tetraphenyl phosphonium chloride :
Molarity = 33.0 mM = 0.033 M (As, 1 mM = 0.001 M)
Volume = 0.45 L
Thus, moles of tetraphenyl phosphonium chloride :
Moles of TPPCl = 0.01485 moles
Molar mass of TPPCl = 342.39 g/mol
The formula for the calculation of moles is shown below:
Thus,
Mass of TPPCl = 5.0845 g
Also,
TPPCl is 94.0 % pure.
It means that 94.0 g is present in 100 g of powder
5.0845 g is present in 5.41 g of the powder.
<u>Answer - 5.41 g</u>