Absorption and emission of electronic energy:
a) You find two bottles in the lab that have lost their labels. You know that the two compounds are 2,3-dimethyl-1,3-butadiene and 2,5-dimethyl-1,3,5-hexatriene. You find that UV absorbance maxima are at $226 \mathrm{~nm}$ and $252 \mathrm{~nm}$. Which spectrum belongs to which compound? How do you know this?
b) The ( $n$ to $\pi^{*}$ ) transition of $\mathrm{N}=\mathrm{O}$ has a maximum at $660 \mathrm{~nm}$, while the ( $\pi$ to $\pi^{*}$ ) transition in $\mathrm{C}=\mathrm{C}$ is at $180 \mathrm{~nm}$. Why would the transition for the $\mathrm{N}=\mathrm{O}$ molecule be at a much longer wavelength? You may wish to use simple HOMO and LUMO energy diagrams for clarity.
c) Diamond and graphite are allotropic forms of carbon. Consider the molecular structures shown and the types of bonds. Use them to explain why graphite (LEFT) is black while diamonds (RIGHT) are colourless.
Segment of a sheet of graphite
Segment of a diamond

Question

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Identify the compounds' structures**: The compound with more conjugated double bonds will have a lower energy gap between the HOMO and LUMO, leading to absorption at a longer wavelength. ‖ ‖ 2 ⋮ **Assign UV spectra**: Since 2,5-dimethyl-1,3,5-hexatriene has three conjugated double bonds, it will absorb at a longer wavelength (252 nm) compared to 2,3-dimethyl-1,3-butadiene, which has two conjugated double bonds and will absorb at 226 nm. ‖ ∻1 Answer∻ ⚹ 2,3-dimethyl-1,3-butadiene absorbs at 226 nm, and 2,5-dimethyl-1,3,5-hexatriene absorbs at 252 nm. ⚹ ∻Key Concept∻ ⚹ Conjugation and UV absorption ⚹ ∻Explanation∻ ⚹ More conjugated double bonds lead to a longer wavelength of UV absorption due to a smaller energy gap between HOMO and LUMO. ⚹ ∻Solution∻ ‖ 1 ⋮ **Compare electronic transitions**: The n to $\pi^*$ transition involves a non-bonding electron (n) moving to an anti-bonding pi orbital ($\pi^*$), while the $\pi$ to $\pi^*$ transition involves an electron from a bonding pi orbital ($\pi$) moving to an anti-bonding pi orbital ($\pi^*$). ‖ ‖ 2 ⋮ **Explain the wavelength difference**: The energy gap between n and $\pi^*$ orbitals is smaller than that between $\pi$ and $\pi^*$ orbitals, resulting in absorption at a longer wavelength for the n to $\pi^*$ transition. ‖ ∻2 Answer∻ ⚹ The n to $\pi^*$ transition for the N=O molecule occurs at a much longer wavelength because the energy gap between n and $\pi^*$ orbitals is smaller than that between $\pi$ and $\pi^*$ orbitals. ⚹ ∻Key Concept∻ ⚹ Energy gap and wavelength ⚹ ∻Explanation∻ ⚹ A smaller energy gap between orbitals results in absorption at a longer wavelength. ⚹ ∻Solution∻ ‖ 1 ⋮ **Analyze the bond types**: Graphite consists of sp2 hybridized carbon atoms forming $\pi$ bonds, while diamond consists of sp3 hybridized carbon atoms with no $\pi$ bonds. ‖ ‖ 2 ⋮ **Explain the color difference**: The $\pi$ bonds in graphite allow for the absorption of visible light, making it appear black. Diamond lacks $\pi$ bonds, so it does not absorb visible light and appears colorless. ‖ ∻3 Answer∻ ⚹ Graphite is black because it absorbs visible light due to the presence of $\pi$ bonds, while diamond is colorless because it lacks $\pi$ bonds and does not absorb visible light. ⚹ ∻Key Concept∻ ⚹ Bond types and light absorption ⚹ ∻Explanation∻ ⚹ The presence of $\pi$ bonds in graphite allows for the absorption of visible light, while the absence of $\pi$ bonds in diamond results in no absorption of visible light. ⚹◊What is the relationship between the molecular structures of diamond and graphite and their differences in color?⍭ Generate me a similar question◊

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