`CH_(3)OC_(2)H_(5)` and `(CH_(3))_(3)COCH_(3)` are treated with hydroiodic acid. The fragments after reaction obtained are

Options

A. `CH_(3)I+HOC_(2)H_(5), (CH_(3))_(3)C-I+HOCH_(3)`
B. `CH_(3)OH+C_(2)H_(5)I, (CH_(3))_(3)Cl+HOCH_(3)`
C. `CH_(3)OH+C_(2)H_(5), (CH_(3))_(3)C-OH+CH_(3)I`
D. `CH_(3)I+HOC_(2)H_(5), CH_(3)I+(CH_(3))_(3)-C-OH`.

Answer

A

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. Hence reaction occurs via `S_(N)2` mechanism and because of the streic effect of the larger group, `I^(-)` attacks the smaller (Me) group.
`CH_(3)OC_(2)H_(5)+HI to CH_(3)I+C_(2)H_(5)OH`
When the substrate is a methyl t-alkyl ether, the products are t-RI and MeOH. Here reaction occurs by `S_(N)I` mechanism and formation of products is controlled by the stability of carbocation. Since carbocation stability order is `3^(@)gt 2^(@)gt 1^(@)gt overset(o+)(C )H_(3)`.
`therefore` Alkyl halide is always derived from tert-alkyl group.
`underset("tert-Butyl methyl ether")(CH_(3)-overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")("C ")-O-CH_(3))+HI overset(373 K)underset(S_(8)l)rarr underset("tert-Butyl iodide")(CH_(3)-overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")("C ")-O-I)+CH_(3)OH`