A) \[C{{l}_{2}}CHCHO\]
B) \[ClC{{H}_{2}}COOH\]
C) \[C{{H}_{3}}COCl\]
D) \[ClC{{H}_{2}}CHO\]
Correct Answer: A
Solution :
\[CH\equiv CH\xrightarrow{HOCl}C{{l}_{2}}CHCHO\]
tautomerising electrophile is \[C\,{{l}^{+}}\] mainly.
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