Deduce the structure of a compound, $\mathrm{C}_4 \mathrm{H}_6 \mathrm{O}$, with the following spectral data: (a) Electronic absorption at $\lambda_{\max }=213 \mathrm{~nm}, \varepsilon_{\max }=7100$ and $\lambda_{\max }=320 \mathrm{~nm}, \varepsilon_{\max }=27$. (b) Infrared bands, among others, at 3000 , 2900,1675 (most intense) and $1602 \mathrm{~cm}^{-1}$. (c) $\mathrm{Nmr}$ singlet at $\delta=2.1 \mathrm{ppm}(3 \mathrm{H}$ s), three multiplets each integrating for $1 \mathrm{H}$ at $\delta=5.0-6.0 \mathrm{ppm}$.
The formula $\mathrm{C}_4 \mathrm{H}_6 \mathrm{O}$ indicates two degrees of unsaturation and may represent an alkyne or some combination of two rings, $\mathrm{C}=\mathrm{C}$ and $\mathrm{C}=\mathrm{O}$ groups.
(a) $\lambda_{\max }$ at $213 \mathrm{~nm}$ comes from the $\pi \rightarrow \pi^*$ transition. It is more intense than the $\lambda_{\max }$ at $320 \mathrm{~nm}$ from the $\mathrm{n} \rightarrow \pi^*$ transition. Both peaks are shifted to higher wavelengths than normal (190 and $280 \mathrm{~nm}$, respectively), thus indicating an $\alpha, \beta$-unsaturated carbonyl compound. The 2 degrees of unsaturation are a $\mathrm{C}=\mathrm{C}$ and a $\mathrm{C}=\mathrm{O}$.
(b) The given peaks and their bonds are $3000 \mathrm{~cm}^{-1}, s p^2 \mathrm{C}-\mathrm{H} ; 2900 \mathrm{~cm}^{-1}, s p^3 \mathrm{C}-\mathrm{H} ; 1675 \mathrm{~cm}^{-1}, \mathrm{C}=\mathrm{O}$ (probably conjugated to $\mathrm{C}=\mathrm{C}$ ); $1602 \mathrm{~cm}^{-1}, \mathrm{C}=\mathrm{C}$. All are stretching vibrations. Absence of a band at $2720 \mathrm{~cm}^{-1}$ means no aldehyde $\mathrm{H}$. The compound is probably a ketone.
(c) The singlet at $\delta=2.1 \mathrm{ppm}$ is from a
<smiles>CC(C)=O</smiles> There are also three nonequivalent vinylic H's $(\delta=5.0-6.0 \mathrm{ppm})$ which intercouple. The compound is
<smiles>C=CC(C)=O</smiles>
shown with nonequivalent H's.