The electric field experienced by the OH group of phenol embedded in the cluster of ammonia molecules depends on the relative orientation of the ammonia molecules, and a critical field of 236 MV cm −1 is essential for the transfer of a proton from phenol to the surrounding ammonia cluster. However, exceptions to this rule were observed, which indicates that the projection of the solvent electric field over the O−H bond is not a definite descriptor of the proton transfer reaction. Therefore, a critical electric field is necessary, but it is not a sufficient condition for the proton abstraction. This, in combination with an adequate solvation of the acceptor ammonia molecule in a triple donor motif that energetically favors the proton transfer process, constitutes necessary and sufficient conditions for the spontaneous proton abstraction. The proton transfer process in phenol-(ammonia) n clusters is statistically favored to occur away from the plane of the phenyl ring and follows a curvilinear path which includes the O−H bond elongation and out-ofplane movement of the proton. Colloquially, this proton transfer can be referred to as a "bend-to-break" process.