Question

CH₃OC₂H₅ and (CH₃)₃ C – OCH₃ are treated with hydriodic acid. The fragments obtained after reactions are
(a) CH₃I + HOC₂H₅; (CH₃)₃CI + HOCH₃
(b) CH₃OH + C₂H₅I; (CH₃)₃Cl + HOCH₃
(c) CH₃OH + C₂H₅I; (CH₃)₃COH + CH₃I
(d) CH₃I + HOC₂H₅; CH₃I + (CH₃)₃ COH.

Answer 1

In CH3OC2H5 1 degree alkyl are attached to ether group on both sides so HI react with compound and I gets add to group which is least stericallly hindered so formation of CH3I and C2H5OH takes places

But on second compound (CH3)3COCH3 there is 3 degree carbon attached to ether group due to which formation of carbonation takes places that is most stable carbonation is formed then I- reacts with carbonation to form alkyl iodide So formation of (CH3)3CI and CH3OH takes place.