MCQ
Chlorine can remove
- ✓$Br$ from $NaBr$ solution
- B$F$ from $NaF$ solution
- C$Cl$ from $NaCl$ solution
- D$F$ from $Ca{F_2}$ solution
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($en=$ ethylenediamine)
Cyclohexane $\xrightarrow{{hv/C{l_2}}}(X)\xrightarrow{{alc.KOH/\Delta }}(Y)\mathop {\xrightarrow{{(i)\,{O_3}}}}\limits_{(ii)\,{H_2}O/Zn} (Z)$
$(Z)$ will be :

$[Figure]$ $\xrightarrow{{{H^ + }/H{g^{2 + }}}}A$
the product '$ A'$ is