Explain how you can distinguish between 1,3-octadiene and 2,4,6-octatriene using UV spectra & why its possible

Answer 1

To clarify: in 1,3-octadiene the electrons of the double bonds are conjugated and delocalized over the four carbons involved in the two double bonds. In 2,4,6-octatriene the electrons of the double bonds are conjugated and delocalized over the six carbons involved in the three double bonds. Since they are delocalized over a longer distance in the octatriene, they are lower in energy and will be at a longer wavelength in the UV spectrum. (Think of any stringed instrument like a guitar, if you make the string shorter the pitch of the sound goes up--higher energy. If you make the string longer, lower energy. The underlying quantum mechanical name for this phenomenon is "particle in a box.")

Answer 2

It has to do with the alkene bonds.

In 1,3, -octadiene you have 2 double bonds that will show up in the can't reminder the regions for double bonds off by hand but its in this region).

Now in 2,4,6,-ocatriene you have 3 double bonds that can undergo a process called hyperconjugation, this will cause the double bonds to form a resonace strucure through out the molecule. This causes a in shift in the peaks of the double bonds to move to the right in the spectra as the enegry level for the double bonds is now at a diffetent wavelength.