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[TRICK] THE DOUBLE BOND EQUIVALENCE CONCEPT

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THE DOUBLE BOND EQUIVALENCE CONCEPT:-


Consider a general organic compound having molecular formula as CaHbNcOdXe ,


where,


C-Carbon,


 


H-Hydrogen,


 


N-Nitrogen,


 


O-Oxygen,


 


X-Halogen.


 


 Using the Double Bond Equivalence we can find the number of the pi bonds present in the compound and in most of the cases also predict the exact structural formula of that compound.

D.B.E = (a+1)-{(b+e-c)/2} [D.B.E :- Double Bond Equivalence]


For Ex.

(1)   D.B.E for CH4 is 0 hence no pi bond is present.


 


(2)   D.B.E for  ethane is 0 hence no pi bond is present.


 


(3)   D.B.E for C2H5O is 0 hence no pi bond is present.

For Various values of D.B.E. what interpretation is to be made is given in the tabular form below:-






Sr.No.




D.B.E values




No. of pi bonds




Possible Configurations




1




1




1




One cyclic ring or one pi bond.




2




2




2




*1 Triple bond(2-pi bonds)




*2 Double Bond(2 pi bond)




*2 Cyclic Rings(2 pi bonds consumed)




*1 cycle & 1 Double bond(1 pi bond consumed in ring       and one in double bond).




3




3




3




*1 Triple bond(contains 2 pi bond)+1 double    bond(contains 1pi bond)




*1 Triple bond(contains 2 pi ond)+1 cycle(1 pi bond consumed)




*1 cycle(1 pi bond consumed)+2 double bond(2 pi bond present)




*3 cycles(contains 3 pi bonds).




4




4




4




One Benzene ring(1 pi bond consumed in the formation of the ring and the other 3 are present in the benzene ring).




5




5




5




In this case break it as 4+1. i.e. 1 Benzene ring +1 Double Bond(4 pi bonds consumed in the benzene ring as described earlier and one pi bond present).




6




6




6




Break as 4+2. i.e. 1 Benzene ring and 2 Double bonds.




7




7




7




Break as 4+3. i.e. 1 Benzene ring and 3 Double bonds.




8




8




8




Break as 4+4. i.e. 2 Benzene rings.



 

If while calculating for D.B.E. the second term in the equation i.e. {(b+e-c)/2} comes out to be fractional then round it up. For  ex. 3.5  to 4, 5.5 to 6. In most of such cases the compound will be a intermediate as described below.


 


While predicting the structures using the above concept sometimes you will be getting   the structures in which the carbon valencies is not satisfied. In such cases the formula may represent  intermediate species such as carbocation , carbanion or carbon free radicals. For Ex.  C7H7 .It represents a Benzyl anion.


 


 Using the concept given above can you predict the structures of compounds given below?


 C12H10, C8H9 (Hint. It represents a cation).

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