Question

Arrange the following compounds in order of increasing dipole moment.
toluene $\left(I\right)$
$m$-dichlorobenzene $\left(II\right)$
$0$-dichlorobenzene $\left(III\right)$
$p$-dichlorobenzene $\left(IV\right)$

• A. $I<IV<II<III$
• B. $IV<I<II<III$
• C. $IV<I<III<II$
• D. $IV<II<I<III$

Answer 1

Answer: (B)

$IV<I<II<III$

The increasing order of dipole moment is $IV<I<II<III$.

$o-,m-,p-$ derivates has $\alpha ={60}^o,{120}^o$ and ${180}^o$ and thus, resultant vector has zero dipole moment in $p$-derivative because of symmetrical structure. Also dipole moment of $m$-dichlorobenzene is more than toluene. $o-$ and $m-$dichlorobenzene have higher dipole moments than toluene due to high electronegativity of chlorine than $C{H}_3$ group. Further, to the $o$-dichlorobenzene has higher dipole moment due to lower bond angle than the $m$-isomer.