Double bond equivalent

The double bond equivalent (DBE) is an aid in determining the structure of an organic molecule if its molecular formula is known. A double bond equivalent corresponds to a double bond or a ring in the molecule. One triple bond corresponds to two double bond equivalents.

calculation

The following formula is used to calculate the double bond equivalent of the empirical formula , where the X stands for halogens : ${\displaystyle \mathrm {C_{c}H_{h}O_{o}N_{n}X_{x}} }$

${\displaystyle \mathrm {DBA} ={\frac {2\ c-h+n-x+2}{2}}}$

It should be noted that the number of oxygen is not included in the formula.

The formula can be extended to sulfur or phosphorus-containing compounds by adding the number of these atoms to the number of oxygen or nitrogen - i.e. atoms of the same valency .

General formula

In the above formula, a system can be seen with regard to the value :

${\displaystyle \mathrm {DBA} ={\frac {2\cdot ({\text{4-wertige Atome}})+({\text{3-wertige Atome}})-({\text{1-wertige Atome}})+2}{2}}}$

Or more generally:

${\displaystyle \mathrm {DBA} =1+{\frac {1}{2}}\sum n_{i}(v_{i}-2)}$,

where n _{i represents} the number of atoms with valence v _i .

Examples

connection	Sum formula	Number of double bonds	Number of rings	Double bond equivalent
benzene	${\displaystyle \mathrm {C_{6}H_{6}} }$	3	1	${\displaystyle {\frac {2\cdot 6-6+2}{2}}=4}$
acetic acid	${\displaystyle \mathrm {C_{2}H_{4}O_{2}} }$	1	0	${\displaystyle {\frac {2\cdot 2-4+2}{2}}=1}$
Ethanol	${\displaystyle \mathrm {C_{2}H_{6}O} }$	0	0	${\displaystyle {\frac {2\cdot 2-6+2}{2}}=0}$
Glycine	${\displaystyle \mathrm {C_{2}H_{5}NO_{2}} }$	1	0	${\displaystyle {\frac {2\cdot 2-5+1+2}{2}}=1}$
Cyclohexanone	${\displaystyle \mathrm {C_{6}H_{10}O} }$	1	1	${\displaystyle {\frac {2\cdot 6-10+2}{2}}=2}$
Ethine	${\displaystyle \mathrm {C_{2}H_{2}} }$	2	0	${\displaystyle {\frac {2\cdot 2-2+2}{2}}=2}$
Caffeine	${\displaystyle \mathrm {C_{8}H_{10}O_{2}N_{4}} }$	4th	2	${\displaystyle {\frac {2\cdot 8-10+4+2}{2}}=6}$
Chloramphenicol	${\displaystyle \mathrm {C_{11}H_{12}Cl_{2}N_{2}O_{5}} }$	5	1	${\displaystyle {\frac {2\cdot 11-12-2+2+2}{2}}=6}$
Spironolactone	${\displaystyle \mathrm {C_{24}H_{32}O_{4}S} }$	4th	5	${\displaystyle {\frac {2\cdot 24-32+2}{2}}=9}$
Cefaclor	${\displaystyle \mathrm {C_{15}H_{14}ClN_{3}O_{4}S} }$	7th	3	${\displaystyle {\frac {2\cdot 15-14-1+3+2}{2}}=10}$

Individual evidence

1. ↑ Jonathan Clayden, Nick Greeves, Stuart G. Warren: Organic Chemistry . 2nd Edition. Springer Spectrum, Berlin Heidelberg 2013, ISBN 978-3-642-34715-3 , p. 83-85 .