A) \[{{C}_{2}}{{H}_{5}}l\]
B) \[C{{H}_{3}}l\]
C) \[CH{{l}_{3}}\]
D) \[C{{H}_{2}}{{l}_{2}}\]
Correct Answer: C
Solution :
Formation of iodoform takes place. \[C{{H}_{3}}C{{H}_{2}}OH+{{l}_{2}}\xrightarrow{\text{Oxidation}}C{{H}_{3}}CHO\xrightarrow{{{I}_{2}}/O{{H}^{-}}}\] \[C{{l}_{3}}CHO\xrightarrow{O{{H}^{-}}}CH{{l}_{3}}+HCO{{O}^{-}}\]You need to login to perform this action.
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