The delocalization energy of 1,3-butadiene

For butadiene we get:

p electron energy of 1,3 butadiene = 2(a+1.62b) + 2(a+0.62b)

= 4a + 4.48b

p electron energy of an equivalent number of isolated double bonds

= 2´(p electron energy of ethene)

For butadiene: p-electron energy of an equivalent number of double bonds

=

2´(p electron energy of ethene)

= 2´(2a+2b)

= 4a+4b

Therefore the delocalization energy of butadiene

= (4a + 4.48b) – (4a+4b)

= 0.48b

Cyclic conjugated polyenes