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Chemistry 14C Course Web Page > Forums > Midterm Exam 1 > DBE
 
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chilicheese21
Registered: 10/23/09
Posts: 3

    11/06/09 at 11:45 PM
  Reply with quote#1

Can someone explain to me how DBE can be calculated and how it really affects the calculation of a molecular structure?
rkeith08
Registered: 10/05/09
Posts: 13

    11/07/09 at 01:08 PM
  Reply with quote#2

DBE's can be calculated using the formula:

# carbons-(# of Hydrogens/Halogens /2)+ (# of nitrogen/2) +1

or C- (H/2) + (N/2) + 1= # of DBE

It affects structure because the number of DBE's allows you to see how many pi bonds or rings are present in the molecule. For example, if you have only 1 DBE you know automatically that there are no benzene rings due to the fact that you don't have 4 DBE's to use (3 for the pi bonds, and 1 for the ring). You know that a DBE has to be a positive number (or else you calculated something wrong) and that 1 DBE can either be used for 1 pi bond, or 1 ring. 2 can either be 1 pi bond, 1 ring or 2 pi bonds, or 2 rings...etc.
DrH
Moderator
Registered: 09/22/08
Posts: 437

    11/09/09 at 07:16 AM
  Reply with quote#3

Lecture supplement, page 102.
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