Question

Write the products will form in the given attachment.....
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Answer 1

1. CH3-O-C2H5 (Excess of HI) --> CH3I + C2H5OH

2. C4H9I + CH3OH

3. C6H5OH + C2H5I

4. C6H5OH + CH3I

Answer 2

products will form :

CH3-O-C2H5 { excess of HI } ---> CH3l + C2H5OH

C4H9l + CH3OH

C6H5OH + C2H5l

C6H5OH + CH3l

hope it helps you dued....♡