Question

1, 3-Butadiene is said to have delocalised bonding. What do you understand by this statement ?​

Answer 1

Answer:

The delocalization energy of 1,3-butadiene

The delocalization energy is the extra stabilization that comes from letting the electrons spread over the whole molecule: each molecular orbital spreads further than just one pair of atoms.

To calculate how much stabilization this provides, in other words to calculate the delocalization energy, we compare the energy of the real molecule with one in which the electrons can’t spread out.

Non-delocalized 1,3-butadiene (with a ‘wall’ between the double bonds):

To find the delocalization energy, first calculate the total p-electron energy of (real) butadiene:

p electron energy of 1,3 butadiene

= 2(a+1.62b) + 2(a+0.62b)

= 4a + 4.48b

Then calculate the energy of the same number of p electrons in isolated (non-delocalized) bonds (i.e. can think of this as butadiene with a ‘wall’ to stop the p electrons spreading):

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

Delocalization energy of allyl (C3H5)

In the allyl radical, there are 3 p electrons.

· 3 carbon 2p orbitals form 3 p molecular orbitals

The energies of the three molecular orbitals are (a + b√2), a and (a – b√2) (bonding, nonbonding and antibonding)

The total p electron energy of allyl radical is therefore:

2´(a + b√2) + a

= 3a + 2b√2

3a + 2.83b

In a model allyl radical with a wall to stop the p electrons delocalizing:

the total p electron energy is

2´(a + b) + a = 3a +2b

(remember a is the energy of an electron in a carbon 2p orbital – you need to add a for the energy of the odd electron)

So the delocalization energy of allyl radical is

3a + 2.83b – 3a +2b

= 0.83b

Have a good day