A) \[C{{H}_{3}}COC{{H}_{3}}\]
B) \[HCHO\]
C) \[C{{H}_{3}}CHO\]
D) \[C{{H}_{3}}C{{H}_{2}}CHO\]
Correct Answer: A
Solution :
The mixture of chloral and caustic soda forms chloroform on distillation. \[\alpha -H\]atom consisting carbonyl compounds form Aldol in the presence of weak base. HCHO does not have\[\alpha \]atom.
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