2-Ethoxy-2,3-dimethylbutane reacts with concentrated aqueous HI to form two initial organic products (A and B). Further reaction
with HI produces organic product C from product B. Draw the structures of these three products.
1 answer:
Answer:
A = 2-iodo-2,3-dimethylbutane
B = Ethanol
C = Iodoethane (also called ethyl-iodide)
Explanation:
2-Ethoxy-2,3-dimethylbutane reacts with conc. HI to cleave the oxy-functional group.
On one end, ethanol is formed and on the other hand, 2-iodo-2,3-dimethylbutane is formed.
But ethanol reacts further with conc HI to give iodoethane.
Therefore,
A = 2-iodo-2,3-dimethylbutane
B = Ethanol
C = Iodoethane (also called ethyl-iodide)
This is all shown in the attached image.
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Answer:
5.758 is the density of the metal ingot in grams per cubic centimeter.
Explanation:
1) Mass of pycnometer = M = 27.60 g
Mass of pycnometer with water ,m= 45.65 g
Density of water at 20 °C = d =
1 kg = 1000 g
Mass of water ,m'= m - M = 45.65 g - 27.60 g =18.05 g
Volume of pycnometer = Volume of water present in it = V
2) Mass of metal , water and pycnometer = 56.83 g
Mass of metal,M' = 9.5 g
Mass of water when metal and water are together ,m''= 56.83 g - M'- M
56.83 g - 9.5 g - 27.60 g = 19.7 g
Volume of water when metal and water are together = v
Density of metal = d'
Volume of metal = v' =
Difference in volume will give volume of metal ingot.
v' = v - V
Since volume cannot be in negative .
Density of the metal =d'
=