Question

Why are the dimethylamine lone pair and the indole lone pair occupying different orbitals? That is, why is the observed arrangement the most favourable in each case? Your answer should make use of VBT concepts (specific orbitals overlapping to form specific bonds, hybridization), but you may also wish to consider alternative Lewis structures for indole. What is the (approximate) hybridization of atomic orbitals at N that accounts for this geometry? Draw an orbital hybridization energy diagram for the N valence electrons and orbitals, and use it to determine in what atomic or hybrid orbital the indole N lone pair formally resides.

Answer 1

In dimethylamine, the hybridization on 'N' is sp³.

Hence, the lone pair of electrons on 'N' in dimethylamine resides in an sp³-hybrid orbital.

Whereas in indole, the hybridization on 'N' is sp².

Hence, the lone pair of electrons on 'N' in indole resides in an sp²-hybrid orbital.

Explanation:

In dimethylamine, an acyclic molecule, i.e. (CH₃)₂NH with a lone pair of electrons stay on 'N' only.

In indole, a cyclic aromatic molecule, i.e. C₈H₆NH with a lone pair of electrons on 'N', but the lone pair of electrons is involved in the aromatic ring current (delocalization) to satisfy the Huckel number of pi-electrons (4n +2, n = 2, i.e. 10 pi-electrons) for aromaticity.