MCQ
The addition of $HBr$ is easiest with
- A$ClC{H_2} = CHCl$
- B$ClCH = CHCl$
- C$C{H_3} - CH = C{H_2}$
- ✓${(C{H_3})_2}C = C{H_2}$
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$(I) HCHO$,
$(II)$ $C{H_3}CHO$,
$(III)$ $C{H_3}COC{H_3}$ is