Final answer:
Benzene absorbs UV light typically near 180 nm to 200 nm due to its conjugated pi-electron system. This absorption is in the UV region, as opposed to the visible region which starts around 400 nm. The absorptions we see are a result of the molecule's π-π* transitions, with n-π* transitions contributing to weaker absorption bands.
Step-by-step explanation:
The question is asking where benzene gives an absorption band in UV spectra. Benzene, which does not have a visible chromophore like beta-carotene, absorbs UV light due to its conjugated pi-electron system. Typically, benzene shows a strong absorption near 180 nm to 200 nm due to π-π* transitions, and weaker absorptions that can extend up to around 250-260 nm because of n-π* transitions. In UV-vis spectroscopy, the electronic transitions of benzene are less energetic than those of isolated double bonds, which occur at shorter wavelengths. The conjugated system of benzene stabilizes the excited state, leading to absorption in the UV region rather than the visible region. Chromophores with extensive conjugation, like in beta-carotene, absorb in the visible range and thus affect the color we perceive. However, benzene's absorption does not extend into the visible spectrum, which typically starts around 400 nm.
Understanding the UV absorption of molecules like benzene is important in organic and biological chemistry, particularly when using spectrophotometric analysis to determine the concentration of substances through the application of the Beer-Lambert Law. The law states that absorbance is directly proportional to concentration within a certain range, which allows for quantitative analysis of compounds with chromophoric groups.