Question

Explain why the $\mathbf{\beta }$ carbon of an ,$\mathbf{\alpha }\mathbf{,}\mathbf{\beta }$-unsaturated carbonyl compound absorbs farther downfield in the ${}^{\mathbf{13}}\mathbf{C}$NMR spectrum than the a carbon, even though the $\mathrm{\alpha }$carbon is closer to the electron-withdrawing carbonyl group. For example, the $\mathbf{\beta }$ carbon of mesityl oxide absorbs at 150.5 ppm, while the carbon absorbs at 122.5 ppm.

Short answer

This is due to the shifting of electrons from the beta carbon towards the carbonyl carbon.

Step-by-step solution

Step 1:  α,β -unsaturated carbonyl

The $\mathrm{\alpha },\mathrm{\beta }$-unsaturated carbonyl compound contains a double bond at the $\mathrm{\alpha }$ and $\mathrm{\beta }$positions. The $\mathrm{\alpha }$position is next to the carbonyl carbon, whereas the $\mathrm{\beta }$ position is next to the $\mathrm{\alpha }$carbon.

The conjugation present between the double bond and the carbonyl group changes the electron density from one place to other. Therefore, the signals may appear downfield and upfield.

Step 2:  C13NMR spectrum

The more electron-dense carbon signals appear upfield, and the less electron-dense carbon atoms signals appear downfield.

Explanation

The double bond (between alpha and beta carbon) and the carbonyl carbon (C=O) are in conjugation. Therefore, the electrons are shifted from the beta carbon towards the carbonyl carbon.

Hence, the beta carbon acquires positive charge, and the alpha carbon is surrounded with more number of electrons.

Hence, the ${}^{13}\mathrm{C}$ NMR signal is shown at 150.5 ppm for beta carbon (downfield) and 122.5 for alpha carbon (upfield).