Question

$CH _{3} OC _{2} H _{5}$ and $\left( CH _{3}\right)_{3} C - OCH _{3}$ are treated with hydroiodic acid. The fragments after reactions obtained are

• A. $CH _{3} I + HOC _{2} H _{5} ;\left( CH _{3}\right)_{3} C -1+ HOCH _{3}$
• B. $CH _{3} OH + C _{2} H _{5} I ;\left( CH _{3}\right)_{3} Cl + HOCH _{3}$
• C. $CH _{3} OH + C _{2} H _{5} ;\left( CH _{3}\right)_{3} C - OH + CH _{3} I$
• D. $CH _{3} I + HOC _{2} H _{5} ; CH _{3} I +\left( CH _{3}\right)_{3}- C - OH$

Answer 1

The Correct Option is A

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 $S _{ N } 2$ mechanism and because of the steric effect of the larger group, $I ^{-}$ attacks the smaller (Me) group. $CH _{3} OC _{2} H _{5}+ HI \rightarrow CH _{3} I + C _{2} H _{5} OH$ When the substrate is a methyl $t$ -alkyl ether, the products are $t-R I$ and MeOH. Here reaction occurs by $S_{N} !$ mechanism and formation of products is controlled by the stability of carbocation. Since carbocation stability order is $3^{\circ}>2^{\circ}>1^{\circ}> \overset{\oplus}{CH _{3}}$ $\therefore$ Alkyl halide is always derived from tert-alkyl group.