A) \[C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}-I+C{{H}_{3}}C{{H}_{2}}OH\]
B) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}+C{{H}_{3}}C{{H}_{2}}OH\]
C) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}OH+C{{H}_{3}}C{{H}_{3}}\]
D) \[C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}OH+C{{H}_{3}}-C{{H}_{2}}-I\]
Correct Answer: D
Solution :
When cone. HI or HBr react with ether, the corresponding alcohol and iodide is formed when there is a case of mixed ethers, the halogen atom attaches to the smaller alkyl group, due to steric effect. \[\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}} \\ & C{{H}_{3}}-\overset{|}{\mathop{C}}\,H-C{{H}_{2}}-O-C{{H}_{2}}-C{{H}_{3}}+HI\xrightarrow[{}]{\Delta } \\ \end{align}\] \[\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}} \\ & C{{H}_{3}}-\overset{|}{\mathop{C}}\,H-C{{H}_{2}}OH+C{{H}_{3}}C{{H}_{2}}I \\ \end{align}\]
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