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.

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Explanation:

In this case, we have the following substract:

Ph-CH2 - O - CH2CH3

That's the benzyl ethyl ether.

When this compound reacts with a concentrated Solution of HI, as we have an acid medium, we will have an SN1 reaction, where the Hydrogen will attach to the oxygen and the iodine will go to the most stable carbon, in this case, the carbon next to the ring is the most stable (because of the double bond of the ring, charges can be stabilized better this way), so product A will be the benzyl with the iodine and product B, will be the alcohol:

A: Ph - CH2 - I

B: CH3CH2OH

When B reacts again with HI, it's promoting another SN1 reaction, where the OH substract the H from the HI, the OH2+ will go out the molecule, leaving a secondary carbocation (CH2+) and then, the Iodine (I-) can go there via SN1 and the final product would be:

C: CH3CH2I

See picture for mechanism: