The S-1<--S0O00 transitions of phenol and the hydrogen bonded phenol(H2O)(1) cluster have been studied by high resolution fluorescence excitation spectroscopy. All lines in the monomer spectrum are split by 56+/-4 MHz due to the internal rotation of the -OH group about the a axis. The barrier for this internal motion is determined in the ground and excited states; V-2 " = 1215 cm(-1), and V-2’ = 710 cm(-1). The rotational constants for the monomer in the ground state are in agreement with those reported in microwave studies. The excited state rotational constants were found to be A’ = 5313.7 MHz, B’ = 2620.5 MHz an C’ = 1756.08 MHz. The region of the redshifted 0(0)(0) transition of phenol(H2O)(1) shows two distinct bands which are 0.85 cm(-1) apart. Their splitting arises from a torsional motion which interchanges the two equivalent H atoms in the (HO)-O-2 moiety of the cluster. This assignment was confirmed by spin statistical considerations. Both bands could be fit to rigid rotor Hamiltonians. Due to the interaction between the overall rotation of the entire cluster and the internal rotation, both bands have different rotational constants. They show that V-2’< V-2 ", and that the internal rotation axis is nearly parallel to the a-axis of the cluster. If it is assumed that the structure of the rotor part does not change upon electronic excitation, the internal motion becomes simply a rotation of the water around it symmetry axis. Assuming this motion, barriers of 180 and 130 cm(-1) could be estimated for the S-0 and S-1 states, respectively. The analysis of the rotational constants of the cluster yielded an O-O distance of the hydrogen bond of 2.93 Angstrom in the ground state and 2.89 Angstrom in the electronically excited state. In the equilibrium structure of the cluster, the plane containing phenol bisects the plane of the water molecule.