"benzyl ethyl ether reacts with concentrated aqueous hi to form two initial organic products (a and b). further reaction of product b with hi produces organic product c. draw the structures of these three products."

See below.  

Step-by-step explanation:

Ethers react with HI at high temperature to produce an alky halide and an alcohol.

R-OR' + HI ⟶ R-I + H-OR'

Benzylic ethers react by an Sₙ1 mechanism by forming the stable benzyl cation.

If there is excess HI, the alcohol formed in Step 2 is also converted to an alkyl iodide:

ROH +HI ⟶ R-I + H-OH

Thus, benzyl ethyl ether reacts to form benzyl iodide (a) and ethanol (b).

The ethanol reacts with excess HI in an Sₙ2 reaction to form ethyl iodide (c).

the answer is 75.0%.

i think its 100.

hope this !