Question

Arrange the following compounds in decreasing order of dipole moment values Explain the order
(a) `CBr_(4)` , (b) `CHBr_(3)` , (c ) `CH_(2)Br_(2)` , (d) `CH_(3)Br` .

Answer 1

The three-dimensional structures of three compounds along with the direction of dipole moments in each of their bonds are shown as below:

I. `Cbr_(4)` is symmetrical and therefore has `mu = 0`.
II. `CHBr_(3)`. the resultant of two `(C - Br)` dipoles, is opposed by the resultant of `(C-H)` and `(C - Br)` bonds, which is smaller than the former. Therefore. `(ChBr_(3))` shows `mu = 1.01D`
III. `CH_(2)Br_(2)`, the resultant of two `(C-Br)` dipole momnets, is reinforced by two `(C-H)` bonds, therefore, `(CH_(2) Br_(2))` shows `mu = 1.52`, which is higher than that of `CHBr_(3)`.
IV. `(CH_(3) - Br)` due to `-I` effect of `Br` has highest `mu. :. mu` order is : `(IV) gt (III) gt (II) gt (I)`.