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
(a) `CBr_(4)` is symmetrical and therefore has `mu =0`
(c ) `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 = 1.01 D`
(c ) `CH_(2)Br_(2)` the resultant of Two `(C -Br)` dipole moments, is reinforced by two `(C -H)` bonds therefore, `(CH_(2)Br_(2)` shows `mu = 1.52` which is higher than that of `CHBr_(3)`
(d) `(CH_(3) -Br)` due to -I effect of br, has the highest mu Therefore, the order of `(mu)` is `(IV) gt (III) gt (II) gt (I)`
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