UV-Visible spectroscopy

Four transition types, ranked by energy. Absorbing a photon in the UV-visible range promotes an electron from a bonding or non-bonding orbital into an antibonding orbital. Highest energy (shortest wavelength) to lowest:

• $\sigma\to\sigma^*$: out of a $\sigma$-bond, needs very high energy ($<150~\text{nm}$, far/vacuum UV), seen in saturated hydrocarbons only.
• $n\to\sigma^*$: a heteroatom lone pair into a $\sigma^*$ orbital, typically $150$-$250~\text{nm}$, seen in alcohols, amines, alkyl halides.
• $\pi\to\pi^*$: out of a $\pi$-bond, needs an unsaturated group (C=C, C=O, aromatic ring); strongest absorptions ($\varepsilon\sim10^3$-$10^5~\text{L mol}^{-1}\text{cm}^{-1}$), and the one that moves into the visible with conjugation.
• $n\to\pi^*$: a lone pair (e.g. carbonyl oxygen) into a $\pi^*$ orbital; weak and symmetry-disfavoured ($\varepsilon\sim10$-$100~\text{L mol}^{-1}\text{cm}^{-1}$), at longer wavelength than the $\pi\to\pi^*$ of the same group.

Chromophores and auxochromes. A chromophore is the functional group that itself absorbs (an isolated C=C or C=O). An auxochrome does not absorb strongly on its own but, attached to a chromophore, donates electron density into the $\pi$ system and shifts/intensifies its absorption (-OH, -NH$_2$, -Cl on a benzene ring or conjugated chromophore).

Conjugation narrows the gap. As more alternating double bonds link together, the $\pi$ electrons delocalize over a longer system. Delocalization lowers the energy of the bonding combinations and raises the energy of the antibonding ones less steeply, so the HOMO-LUMO gap shrinks and $\lambda_{max}$ moves to longer wavelength. Four standard shift terms describe this:

• Bathochromic (red) shift — $\lambda_{max}$ to longer wavelength.
• Hypsochromic (blue) shift — $\lambda_{max}$ to shorter wavelength.
• Hyperchromic effect — absorption intensity ($\varepsilon$) increases.
• Hypochromic effect — absorption intensity ($\varepsilon$) decreases.

Quantitatively, once a chromophore is chosen, the absorbance measured at its $\lambda_{max}$ follows the Beer-Lambert law:

where: $A$ is absorbance (dimensionless), $\varepsilon$ is the molar absorptivity ($\text{L mol}^{-1}\text{cm}^{-1}$, characteristic of the transition), $c$ is molar concentration ($\text{mol L}^{-1}$), and $l$ is the path length through the sample (cm, usually $1.00~\text{cm}$ for a standard cuvette).

Problem: A dye solution has $\lambda_{max}=480~\text{nm}$ with molar absorptivity $\varepsilon = 2.10\times10^4~\text{L mol}^{-1}\text{cm}^{-1}$. In a $1.00~\text{cm}$ cuvette at $480~\text{nm}$, a sample of unknown concentration gives $A=0.630$. Find the concentration, identify the likely transition type, and then find what path length would be needed to read exactly $A=1.000$ for the same solution.

Step 1. Rearrange Beer-Lambert for concentration: $c = \dfrac{A}{\varepsilon l} = \dfrac{0.630}{(2.10\times10^4~\text{L mol}^{-1}\text{cm}^{-1})(1.00~\text{cm})} = 3.00\times10^{-5}~\text{mol L}^{-1}$.

Step 2. Identify the transition: $\lambda_{max}=480~\text{nm}$ is in the visible region, and $\varepsilon\approx2\times10^4~\text{L mol}^{-1}\text{cm}^{-1}$ is large, the signature of a fully allowed transition. This is a $\pi\to\pi^*$ transition in an extensively conjugated chromophore (typical of azo dyes) rather than a weak $n\to\pi^*$ transition, which would give $\varepsilon$ two to three orders of magnitude smaller.

Step 3. For the follow-up, hold $\varepsilon$ and $c$ fixed (same solution, same wavelength) and solve for $l$: $l=\dfrac{A}{\varepsilon c}=\dfrac{1.000}{(2.10\times10^4)(3.00\times10^{-5})}=\dfrac{1.000}{0.630}\approx1.587~\text{cm}$. The concentration is $3.00\times10^{-5}~\text{mol L}^{-1}$, the transition is a conjugated $\pi\to\pi^*$, and a $1.59~\text{cm}$ cuvette would give exactly $A=1.000$.

Competitive stretch

Unconjugated alkenes absorb near $\lambda_{max}\approx175~\text{nm}$ ($\pi\to\pi^*$); 1,3-butadiene, with two conjugated double bonds, absorbs at $\lambda_{max}\approx217~\text{nm}$. In the delocalized picture, the four $2p$ orbitals of the four sp² carbons combine into four $\pi$ molecular orbitals ($\pi_1,\pi_2,\pi_3^*,\pi_4^*$); the filled HOMO is $\pi_2$ and the empty LUMO is $\pi_3^*$. Delocalization narrows this HOMO-LUMO gap below the localized $\pi$-$\pi^*$ gap of an isolated double bond, giving the observed bathochromic shift from $175$ to $217~\text{nm}$ — the same mechanism, extended over roughly eleven conjugated double bonds, that makes carotenoids strongly coloured.

• Assuming "colourless" means "no UV-Vis absorption." Acetone is colourless yet absorbs strongly in the UV (its carbonyl transitions sit below about $350~\text{nm}$, outside the visible range the eye detects); UV-Vis absorption and visible colour are not synonyms.
• Mixing up which transition is intense. $n\to\pi^*$ is weak (symmetry-disfavoured) and $\pi\to\pi^*$ is strong (allowed); do not assume all electronic transitions give comparably sized peaks.
• Calling an auxochrome a chromophore. An auxochrome (-OH, -NH$_2$) has no strong absorption of its own; it only modifies an existing chromophore's absorption once attached to it.
• Trusting Beer-Lambert at high concentration. The linear $A=\varepsilon cl$ relationship holds only in the dilute regime; aggregation and intermolecular interactions cause real deviations at high $c$.

Ask the tutor: if you want more practice converting between conjugation length and lambda max, or extra Beer-Lambert rearrangement problems, just ask.