UV/Vis spectroscopy | Spectroscopy | Organic chemistry | Khan Academy

Explain that different molecules absorb different wavelengths of light in the UV and visible regions (200-800 nm). Demonstrate how UV/Vis spectrophotometry reveals which wavelengths a compound absorbs by producing an absorption spectrum.

Examine the absorption spectrum of 1,3-butadiene and identify the lambda max (λmax) at approximately 217 nm as the wavelength of strongest absorption. Note that this UV absorption means butadiene is colorless.

Describe how four sp² hybridized carbons in butadiene create four p orbitals that combine into four molecular orbitals: two bonding and two antibonding. Fill the four π electrons into the lowest energy orbitals in the ground state.

Identify the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) in butadiene. Explain that light absorption excites a π electron from HOMO to LUMO, requiring a specific energy corresponding to the wavelength difference.

Derive the relationship between photon energy (E = hν), frequency, and wavelength, showing that energy is inversely proportional to wavelength. Apply this to butadiene's 217 nm absorption corresponding to the HOMO-LUMO energy gap.

Analyze ethanal's structure showing two π electrons in a C=O bond. Demonstrate the π-π* (pi-star) transition occurring at approximately 180 nm, representing excitation between the bonding and antibonding π orbitals.

Show that oxygen's lone pair electrons occupy a non-bonding orbital (n) with energy between the π bonding and π antibonding orbitals. Explain that the n-π* transition (approximately 290 nm) involves smaller energy gap than π-π*, resulting in longer wavelength absorption.