Question

Consider the following haloalkanes
$\left(\right.1\left.\right)$ $CH_{3}F$
$\left(\right.2\left.\right)$ $CH_{3}Cl$
$\left(\right.3\left.\right)$ $CH_{3}Br$
$\left(\right.4\left.\right)$ $CH_{3}I$
The correct sequence of increasing order of dipole moments is

• A. 1 < 2 < 3 < 4
• B. 4 < 3 < 2 < 1
• C. 4 < 3 < 1 < 2
• D. 3 < 4 < 1 < 2

Answer 1

Answer: (C)

$\underset{1.64 D}{C H_{3} I} < \underset{1.79 D}{C H_{3} B r} < \underset{1.84 D}{C H_{3} F} < \underset{1.94 D}{C H_{3} C l}$
The charge separation in $C-X$ bond decreases in the order
$C-F>C-Cl>C-Br>C-I,$ due to decrease in the electronegativity as we more from $F \rightarrow Cl \rightarrow Br \rightarrow I$ .
$CH_{3}Cl$ has higher dipole moment than $CH_{3}F$ , because of larger bond length of $C-Cl$ bond. The larger $C-Cl$ bond length causes dipole moment $\left(\right.\mu \left.\right)=q\times d$ to be larger for $CH_{3}Cl$ than $CH_{3}F$ . In case of other haloalkanes, the effect of larger $C-X$ bond length cannot outweigh the effect of decreased charge separation.