A) Ethanal
B) Ethanol
C) Ethanoic acid
D) Ethene
Correct Answer: B
Solution :
\[A\xrightarrow[[O]]{N{{a}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}_{2}}S{{O}_{4}}}\,B\,\,\xrightarrow[{}]{{{H}_{2}}/Ni}\]Ethyl alcohol Compound B on reduction with\[{{H}_{2}}/\]in presence of Ni given ethyl alcohol it means compound B is ethanal, because aldehyde on reduction gives primary alcohol. Again compound B obtained from oxidation of A it means A is ethanol, because primary alcohol on oxidation gives aldehyde. \[\underset{(A)}{\mathop{C{{H}_{3}}-C{{H}_{2}}-OH}}\,\xrightarrow[[O]]{N{{a}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}_{2}}S{{O}_{4}}}\] \[\underset{(B)}{\mathop{C{{H}_{3}}-\overset{O}{\mathop{\overset{|\,|}{\mathop{C}}\,}}\,-H}}\,\xrightarrow{{{H}_{2}}/Ni}\underset{ethyl\,\,alcohol\,\,or\,\,ethanol}{\mathop{C{{H}_{3}}-C{{H}_{2}}-OH}}\,\]You need to login to perform this action.
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