Question: Provide the following information about L-dopa.

a. Label all ${\mathbf{sp}}^{\mathbf{3}}$hybridized C atoms.

b. Label all H atoms that bear a partial positive chargelocalid="1648291738174" $\left({\delta }^{+}\right)$.

c. Draw another resonance structure.

When one s (valence) and three p (valence) orbitals hybridize, they form a hybrid orbital called ${\mathbf{sp}}^{\mathbf{3}}$.

A dipole is created when there is the formation of ${\mathrm{\delta }}^{+}\mathrm{and}{\mathrm{\delta }}^{-}$charges on the atoms involved in the bond.

In the dipole, role="math" localid="1648290916861" ${\mathrm{\delta }}^{+}$is assigned to the less electronegative atom, while ${\mathrm{\delta }}^{-}$is assigned to the more electronegative atom.

The movement of pi electrons generates resonance structures. So, if the structure contains two alternate pi bonds, it can form a resonance structure by changing the pi electron’s place.

a. Carbon has four valence electrons of which two are in the 2s orbital and the other two are in the 2p orbital. It hybridizes by transferring one electron from 2s to 2p, and hence enables ${\mathrm{sp}}^3$hybridization.

Carbon can form a maximum of 4 single bonds through this hybridization. Therefore, L-dopa contains only two${\mathrm{sp}}^3$hybridized carbon atoms as they form four single bonds.

${\mathrm{sp}}^3$hybridized carbon atoms

b. The partial positive charge$\left({\mathrm{\delta }}^{+}\right)$generates upon hydrogen when it bonds with a more electronegative atom (O, N).

The highly electronegative atom attracts the shared pair of electrons towards itself and makes hydrogen partially electron deficient.

Therefore, the partial positive charge$\left({\mathrm{\delta }}^{+}\right)$on hydrogen is shown adjacent to the most electronegative atoms, that is, nitrogen and oxygen which is shown below:

Representation of charges

c. The resonance structure of L-dopa is formed by the movement of pi electrons in the benzene ring and the acid group, as shown below:

Resonance structure