A) \[N{{a}_{2}}C{{r}_{2}}{{O}_{7}}\]and\[NaOH\]
B) \[N{{a}_{2}}C{{r}_{2}}{{O}_{7}}\]and\[dil\,\,{{H}_{2}}S{{O}_{4}}\]
C) \[NaOH\]
D) \[Fe\]in presence of\[NaOH\]
Correct Answer: B
Solution :
\[C{{H}_{3}}C{{H}_{2}}OH\]converts into \[C{{H}_{3}}CHO\]in the presence of \[N{{a}_{2}}C{{r}_{2}}{{O}_{7}}\] and \[dil\,\,{{H}_{2}}S{{O}_{4}}\] on oxidation \[\underset{Ethyl\,\,alcohol}{\mathop{C{{H}_{3}}C{{H}_{2}}OH+(O)}}\,\xrightarrow{N{{a}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}_{2}}S{{O}_{4}}(dil)}\]\[\underset{Acetaldehyde}{\mathop{C{{H}_{3}}CHO+{{H}_{2}}O}}\,\]
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