Question

CH₃OC₂H₅ and (CH₃)₃ C - OCH₃ are treated with hydroiodic acid. The fragments after reactions obtained are

(a) CH₃I + HOC₂H₅; (CH₃)₃C - I + HOCH₃

(b) CH₃OH + C₂H₅I; (CH₃)₃Cl + HOCH₃

(c) CH₃OH + C₂H₅; (CH3)₃C - OH + CH₃I

(d) CH₃I + HOC₂H₅; CH₃I + (CH₃)₃C - OH

Answer 1

The correct option is: (a) CH₃I + HOC₂H₅; (CH₃)₃C - I + HOCH₃

Explanation:

When mixed ethers are used, the alkyl iodide produced depends on the nature of alkyl groups. If one group is Me and the other a pri- or sec-alkyl group, then methyl iodide is produced. Here reaction occurs via S_N2 mechanism and because of the steric effect of the larger group, I^- attacks the smaller (Me) group.

CH₃OC₂H₅ + HI → CH₃I + C₂H₅OH

When the substrate is a methyl t-alkyl ether, the products are t-RI and MeOH. Here reaction occurs via S_N1 mechanism and formation of products is controlled by the stability of carbocation. Since carbocation stability order is 

alkyl halide is always derived from tert-alky| group.