- AC−F
- BC−Cl
- CC−I
- DC−Br
Explanation:
Down the group electronegativity decreases.
C−I It has the least electronegativity difference among carbon-halogen bond.
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$(p)$ $\begin{matrix}
Ph\,\,\,\,\,O\, \\
|\,\,\,\,\,\,\,\,||\,\, \\
Ph-C-C-C{{H}_{3}} \\
|\,\,\,\,\,\,\,\,\,\,\,\, \\
C{{H}_{3}}\,\,\,\,\,\,\,\, \\
\end{matrix}$ $(q)$ $\begin{matrix}
\,Ph\,\,\,\,\,O\, \\
\,\,|\,\,\,\,\,\,\,\,|| \\
Ph-C-C-Et \\
|\,\,\,\,\,\,\,\,\,\, \\
Et\,\,\,\,\,\,\,\,\,\, \\
\end{matrix}$
$(r)$ $\begin{matrix}
\,Ph\,\,\,\,\,O\, \\
|\,\,\,\,\,\,\,\,\,\,||\, \\
Ph-C-C-C{{H}_{3}} \\
|\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
Et\,\,\,\,\,\,\,\,\,\,\,\,\, \\
\end{matrix}$ $(s)$ $\begin{matrix}
\,\,\,Ph\,\,\,\,\,O\, \\
\,\,|\,\,\,\,\,\,\,\,|| \\
Ph-C-C-Et \\
|\,\,\,\,\,\,\,\,\,\, \\
C{{H}_{3}}\,\,\,\, \\
\end{matrix}$
When $(A)$ and $(B)$ reacts with $H_2SO_4$ products obtained are
($en=$ ethylenediamine)
