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.

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.

Hope this Helps!!!

This is what i found on google “when one phosphate group is removed by breaking a phosphoanhydride bond in a process called hydrolysis, energy is released, and atp is converted to adenosine diphosphate (adp). likewise, energy is also released when a phosphate is removed from adp to form adenosine monophosphate (amp).” i hope this you .

The answer is letter c