Question

analyse UV and IR spectra of following comp. HC=CH , H2C=CH2 , H3C-CH3 ​

Answer 1

Answer:

Diene: molecule with two double bonds

Conjugated diene: alternating double and single bonds

When the carbons of a conjugate diene all lie in the same plane, the

p-molecular orbitals overlap.

p-molecular orbitals of butadiene

0 Nodes

3 bonding interactions

0 antibonding interactions

BONDING MO

1 Nodes

2 bonding interactions

1 antibonding interactions

BONDING MO

2 Nodes

1 bonding interactions

2 antibonding interactions

ANTIBONDING MO

3 Nodes

0 bonding interactions

3 antibonding interactions

ANTIBONDING MO

y2 is the Highest Occupied Molecular Orbital (HOMO)

y3 is the Lowest Unoccupied Molecular Orbital (LUMO)

y1 of Butadiene (no nodes, bonding MO)

Bonding Interaction

The four p-electrons in y1 are delocalized

over the four p-orbitals

• • • •

_

+

• • • _

+

•

H3C CH3 H2C CH2 H2C CH CH3 H2C CH CH CH2 H2C CH CH CH2

Table 14.2 (p. 528): Bond lengths in pm

154 133 149 148 134

Explanation: