Question

$(C{H}_3{)}_{3}COC{H}_3$ and $C{H}_{3}O{C}_2{H}_5$ are treated with hydroiodic acid.The fragments obtained after reactions are respectively:

• A. $(C{H}_3{)}_{3}CI+C{H}_{3}OH;C{H}_{3}I+C_2{H}_{5}OH$
• B. $(C{H}_3{)}_{3}CI+C{H}_{3}OH;C{H}_{3}OH+C_2{H}_{5}I$
• C. $(C{H}_3{)}_{3}COH+C{H}_{3}I;C{H}_{3}OH+C_2{H}_{5}I$
• D. $C{H}_{3}I+(C{H}_3{)}_{3}COH;C{H}_{3}I+C_2{H}_{5}OH$

Answer 1

The correct option is A $(C{H}_3{)}_{3}CI+C{H}_{3}OH;C{H}_{3}I+C_2{H}_{5}OH$

When mixed ethers are used, the formation of alkyl iodide depends on the nature of alkyl groups. Methyl iodide is formed when one group is methyl and the other a primary or secondary alkyl group. Here reaction follows $S_N{}^2$ mechanism and because of the steric effect of the larger group, $I^{-}$ attacks the smaller (methyl) group.

$C{H}_{3}O{C}_2{H}_5+HI\to C{H}_{3}I+C_2{H}_{5}OH$
When the substrate is a methyl t-alkyl ether, the products are tertiary alkyl iodide and methanol. Here reaction follows $S_N^1$ mechanism and formation of products is controlled by the stability of carbocation. Since carbocation stability order is $3^0>2^0>1^0>C{H}_3$, therefore alkyl halide is always derived from tert-alkyl group.

$C{H}_3-\stackrel{C{H}_3}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-O-C{H}_3+HI\underset{S_{N}I}{\overset{373K}{\to }}C{H}_3-\stackrel{C{H}_3}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-I+C{H}_{3}OH$
$\text{tert-Butyl methyl ether}$ $\text{tert-Butyl iodide}$