Explain why:
$(i)$ the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
$(ii)$ alkyl halides,though polar,are immiscible with water?
$(iii)$ Grignard reagents should be prepared under anhydrous conditions?

(N/A) $(i)$ In chlorobenzene,the $Cl$ atom is linked to a $sp^{2}$ hybridized carbon atom. In cyclohexyl chloride,the $Cl$ atom is linked to a $sp^{3}$ hybridized carbon atom. The $sp^{2}$ hybridized carbon has more $s$-character than the $sp^{3}$ hybridized carbon atom,making it more electronegative. Consequently,the electron density of the $C-Cl$ bond near the $Cl$ atom is less in chlorobenzene than in cyclohexyl chloride. Furthermore,the $-I$ effect of the benzene ring in chlorobenzene decreases the electron density of the $C-Cl$ bond near the $Cl$ atom. As a result,the polarity of the $C-Cl$ bond in chlorobenzene decreases,leading to a lower dipole moment compared to cyclohexyl chloride.
$(ii)$ To be miscible with water,the solute-water force of attraction must be stronger than the solute-solute and water-water forces of attraction. Alkyl halides are polar molecules held together by dipole-dipole interactions,while water molecules are held together by strong hydrogen bonds. The new force of attraction between alkyl halides and water molecules is weaker than the existing alkyl halide-alkyl halide and water-water forces of attraction. Hence,alkyl halides are immiscible with water.
$(iii)$ Grignard reagents $(R-Mg-X)$ are highly reactive. In the presence of moisture,they react with water to form alkanes:
$R-Mg-X + H_{2}O \rightarrow R-H + Mg(OH)X$
Therefore,Grignard reagents must be prepared under anhydrous conditions to prevent decomposition.