The compound shown is treated with $\mathrm{NaNH}{}_2$ in liquid ammonia and reacts through benzyne intermediates. Considering every benzyne that can form by removal of a hydrogen ortho to chlorine, and both positions at which amide can add to each benzyne, how many different isomeric methylanilines (toluidines) can be formed? Answer as an integer.

The compound is 3-chlorotoluene (chlorine at C-1, methyl at C-3). Removing H at C-2 gives the C1-C2 benzyne: amide at C-1 gives 3-methylaniline, at C-2 gives 2-methylaniline. Removing H at C-6 gives the C1-C6 benzyne: amide at C-1 gives 3-methylaniline again, at C-6 gives 4-methylaniline. The distinct products are o-, m- and p-toluidine: 3 (RDKit canonical SMILES give 3 distinct compounds).