A) chloral
B) \[C{{H}_{3}}Cl\]
C) \[C{{H}_{3}}Cl\]
D) chloroacetic acid
Correct Answer: A
Solution :
Chloral is obtained by the action of chlorine c ethyl alcohol (i) \[C{{l}_{2}}+\underset{ethyl\,alcohol}{\mathop{C{{H}_{4}}C{{H}_{2}}OH}}\,\xrightarrow{{}}\underset{acetaldehyde}{\mathop{C{{H}_{3}}CHO}}\,+2HCl\] \[C{{H}_{3}}CHO+3C{{l}_{2}}\xrightarrow{{}}\underset{chloral}{\mathop{CC{{l}_{3}}CHO}}\,+3HCl\] Chloral reduces Fehling solution and on oxidation gives a trichloro ethanoic add (monocarboxylic acid). \[\underset{(A)}{\mathop{\underset{chloral}{\mathop{CC{{l}_{3}}CHO}}\,}}\,+2CuO\xrightarrow{{}}CC{{l}_{3}}\underset{(B)}{\mathop{-COOH}}\,\]\[\underset{(red\,\,ppt.)}{\mathop{\underset{cuprous\,oxide}{\mathop{+C{{u}_{2}}O}}\,}}\,\] So, compound A is chloral \[(CC{{l}_{3}}CHO)\]
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