Compound W, C6H11Cl, undergoes base-promoted E2 elimination to form a single alkene Y. Under similar reaction conditions X, C6H1
1Br, affords Y and an isomeric alkene. Catalytic hydrogenation of Y produces methylcyclopentane. W readily undergoes an SN2 reaction with sodium iodide in acetone. X does not undergo a similar SN2 reaction. What are the structures of W and X
1 answer:
Answer:
; The structures of both compounds are shown in the attached image.
Explanation:
First, compound W has the molecular formula C6H11Cl, it is observed that it does not have double bonds. removal of E2 produces compound Y, which is an alkene, which is hydrogenated and produces a methylcyclopentane.
Compound W is an alkyl halide that undergoes an SN1 reaction with sodium iodide in the presence of acetone forming the corresponding iodine compound.
while compound X is a tertiary alkyl halide, it is sterically restricted to undergo SN2 reactions.
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Answer:- 47.62 mL
Solution:- It is a dilution problem where we are asked to calculate the volume of 15.75 M perchloric acid solution required to make 500.0 mL of 1.500 M solution.
For solving this type of problems we use the dilution equation:
Where, is the concentration of the concentrated solution and
is it's volume.
is the concentration of the diluted solution and
is it's volume. Let's plug in the values in the equation and solve it for
.
On rearranging this for :
So, 47.62 mL of 15.75 M perchloric acid are required to make 500.0mL of 1.500 M solution.