Arrange the following compounds in order of increasing dipole moment : (I) Toluene (II) $m$ -dichlorobenzene…

Arrange the following compounds in order of increasing dipole moment :
(I) Toluene
(II) $m$ -dichlorobenzene
(III) $o$ -dichlorobenzene
(IV) $p$ -dichlorobenzene
  1. $\quad \mathrm{I} < \mathrm{IV} < \mathrm{II} < \mathrm{III}$
  2. $\mathrm{IV} < \mathrm{I} < \mathrm{II} < \mathrm{III}$
  3. $\quad$ IV $ < $ I $ < $ III $ < $ II
  4. $\quad \mathrm{IV} < \mathrm{II} < \mathrm{I} < \mathrm{III}$

Solution

In $p$ -dichlorobenzene, the two equal dipoles are in opposite direction, hence the molecule has zero dipole moment. In $o$ - and $m$ - dichlorobenzenes, the two dipoles are at $60^{\circ}$ and $120^{\circ}$ apart respectively, and thus according to parallelogram law of forces, the dipole moment of $o$ -dichlorobenzene is much higher than that of $m$ -isomer. Lastly, toluene with a +I group possesses little dipole moment. Thus the overall order is
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Asked in: JEE-TOPICTESTS-CHEMISTRY

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