Question #cc296

1 Answer
Prachi D.
Feb 16, 2018

Following is my approach.

Explanation:

In ‎Chloroform (CHCl3) ,
4 Sigma bonds are present in following way -
one bond b/w C - H and three bond b/w C - Cl
In Acetone (CH3COCH3) ,
9 Sigma bonds and 1 pi bond is present in following way -
one pi bond b/w C = O , 6 sigma bond b/w C - H and 2 sigma bond b/w C - C

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