MCQ on UV-Visible spectroscopy: Page-2

The given structure is 1,3-pentadiene and since it has pi bonds it can undergo pi to pi transition. Therefore the diene acts as chromophore in this molecule with a base value of 217 nm.

First let’s give numbering to the chromophore. Here diene is the chromophore which has a base value as 217 nm. Now let’s see any groups attached to this chromophore. One methyl group is present at 4th position, so we have to give an increment of 5 nm. Combining these, the total value for lambda max is 217 + 5=222 nm

Water has sigma and n electrons. So it can show sigma to sigma and n to sigma transitions. These two types of transitions fall under vacuum UV hence water doesn’t absorb radiation within the UV-Visible region.

The energy required for a transition depends on the energy gap between the bonding molecular orbital and anti-bonding molecular orbital of the particular electron. For example, the energy gap is highest between sigma and sigma* requiring more energy for this transition. Similarly, the energy gap is least between n electrons and pi*.

Note that electrons that form either sigma or pi bond only will have anti-bonding molecular orbital. n electrons are not involved in bonding hence don’t have excited state.

Since the chromophore is benzaldehyde, the base value will be 250 nm according to Woodward-Fieser rules.