LnCl 3 (LnNd, Er) reacts with K 2C 8H 8 to yield the complex (C 8H 8)LnCl·2THF, which reacts with K(2,4-C 7H 11) (2,4-C 7H 11 = 2,4-dimethylpentadienyl) to form (C 8H 8)Ln(2,4-C 7H 11)·THF. The compound (C 8H 8)Nd(2,4-C 7H 11). THF(1) crystallizes from the mixed solvents hexane and THF in monoclinic space group P2 1, with a = 8.734(1), b = 10.915(4), c = 9.786(2) Å, β = 104.13(1)°, V= 904.8(4) Å 3, D c = 1.53 g cm −3 and Z = 2. The crystal of (C 8H 8)Er(2,4-C 7H 11). THF ( 2) belongs to monoclinic space group P21/ m with a = 7.776(2), b = 13.030(5), c = 8.729(3) Å, β = 101.21(2)°, V = 867.59(0.46) Å 3, D c = 1.73 g cm −3 and Z = 2. Because of the difference between the radii of Nd and Er, 2 is not isostructural with 1. The THF molecule in 2 is situated against the open jaws of the pentadienyl ligand, and in compound 1, it is situated by the back of the pentadienyl ligand.