# Why is 1H-borepine aromatic?

For a compound to be aromatic, each atom in the ring must be a part of conjugation. Boron here has no lone pair, so how can 1⁠H-borepine be aromatic as I was told by someone?

1H-borepine is isoelectronic to the cycloheptatrienyl cation. In both cases, you have a planar ring of seven atoms, whose hybridisation is close enough to $\mathrm{sp^2}$ to allow for a third p-orbital to be perpendicular to the ring plane. And in both, you are then filling these seven π orbitals with 6 π electrons, resulting in a Hückel-aromatic system ($4n+2 = 6$ for $n=1$).
• We can make contributing structures in cyclopropenone where a $p$ orbital on carbon is formally empty, the carbon is positively charged (cyclopropenyl cation), and the negative charge is on a substituent (oxygen). Such an aromatic zwitterion structure is not really accurate, but cyclopropenone shows properties consistent with a zwitterionic contribution (basicity, dipole moment). – Oscar Lanzi Oct 25 '17 at 16:49