A) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{3}}+C{{H}_{3}}C{{H}_{2}}OH\]
B) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{2}}OH+C{{H}_{3}}C{{H}_{3}}\]
C) \[C{{H}_{3}}-\underset{{}}{\mathop{\underset{{}}{\mathop{\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,H-C{{H}_{2}}OH+C{{H}_{3}}-C{{H}_{2}}-I\]
D) \[C{{H}_{3}}-\underset{{}}{\mathop{\underset{{}}{\mathop{\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,H-C{{H}_{2}}-I+C{{H}_{3}}C{{H}_{2}}OH\]
Correct Answer: C
Solution :
[c] In the cleavage of mixed ethers having two different alkyl groups, the alcohol and alkyl iodide formed depends on the nature of alkyl groups. When primary or secondary alkyl groups are present, it is the lower alkyl group that forms alkyl iodide therefore \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}-O-C{{H}_{2}}-C{{H}_{3}}+HI\xrightarrow{\Delta }C{{H}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}OH+C{{H}_{3}}C{{H}_{2}}I\]
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