A large isotopic-spin violation was observed in the Ne20(d, α)F18 reaction. The yield to the second T=1 level (3.063 MeV) in F18 was measured and compared with the yield to an adjacent (3.133 MeV) T=0 level. The ratio of the yields to these two levels, averaged over angles from 50° to 140°, for an input energy of 4.00 MeV, was found to be 0.40. A value of approximately 1 was found for this ratio using the O16(He, p)F18 reaction. The isotopic-spin selection-rule breakdown is attributed to Coulomb mixing of states in the compound nucleus Na22 and in the residual nucleus F18. Angular distributions of 11 proton groups from the O16(He3, p)F18 reaction and of seven alpha groups from the Ne20(d, α)F18 reaction are presented.