MCQ
$C{H_2} = CHCl$ reacts with $HCl$ to form
- A$C{H_2}Cl - C{H_2}Cl$
- ✓$C{H_3} - CHC{l_2}$
- C$C{H_2} = CHCl.HCl$
- DNone of these

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In the above reaction $A, B$ and $C $ are the following compounds
[Given $pK _{ a }$ of $CH _{3} COOH =4.57$ ]
