Question

The reaction of $C{H}_{3}O{C}_2{H}_5$ with $HI$ gives:

• A. $C{H}_{3}I$
• B. $C_2{H}_{5}OH$
• C. $C{H}_{3}I+C_2{H}_{5}OH$
• D. $C{H}_{3}OH+C_2{H}_{5}I$

Answer 1

The correct option is C $C{H}_{3}I+C_2{H}_{5}OH$

$C{H}_{3}O{C}_2{H}_5+HI\to C{H}_{3}I+C_2{H}_{5}OH$
In the case of mixed ethers such as methyl ethyl ether, the halogen attaches to simpler alkyl group. So the products are $C{H}_{3}I$ and $C_2{H}_{5}OH$, but not the $C_2{H}_{5}I$ and $C{H}_{3}OH$.