Question

Arrange the following compounds in decreasing order of dipole moment values.
A) CBr4​
B) CHBr3CHBr3
C) CH2Br2CH2Br2
D) CH3Br

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Answer 1

Explanation:

sorry I don't do this topic In my class

Answer 2

Answer:

D>C>B>A That is  CH₃Br> CH₂Br₂>CHBr₃>CBr₄

Explanation:

A) CBr₄ is symmetrical and so it has μ=0

B) CHBr₃ is 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 (CHBr₃) shows μ=1.01D

C) CH₂Br₂ is the resultant of two (C-Br) dipole moment is reinforced by two (C-H) bonds. therefore, ( CH₂Br₂) shows  μ=1.52, which is higher than that of CHBr₃.

D) (CH₃-Br) due to -I effect of Br has highest   μ. therefore   μ order is D>C>B>A

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