MCQ
Oxidising action increases in halogen in the following order
- A$Cl < Br < I < F$
- B$Cl < I < Br < F$
- C$I < F < Cl < Br$
- ✓$I < Br < Cl < F$
Among halogens $F$ is the directly most powerful oxidising agent.
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Product $(A)$ of the reaction is
$n-$propyl bromide $\xrightarrow{{Na}}(A)\xrightarrow[{A{l_2}{O_3}/773\,K}]{{Cr{O_3}}}(B)\xrightarrow[{500\,K}]{{C{l_2}/UV}}(C)$
$n-$propyl bromide $\xrightarrow{{Na}}(A)\xrightarrow[{A{l_2}{O_3}/773\,K}]{{Cr{O_3}}}(B)$ $\xrightarrow[{}]{{C{H_3} - Cl/AlC{L_3}}}(D)\xrightarrow{{C{l_2}/hv}}(E)$