Two-step Synthesis of 9-Fluorenemethanol from Fluorene

Using fluorene as raw material, under alkali catalysis, fluorene is first converted into 9-fluorene carboxaldehyde, and then reduced to 9-fluorene methanol. The effects of acylation reagents, reducing agents, types of auxiliary agents, ratio of reactants, temperature and time on the reaction were investigated. The results show that fluorene can be efficiently converted into 9- Fluorene formaldehyde; no need to change the solvent in the reduction stage, using NaBH4 as reducing agent, adding soluble inorganic metal salt additives, such as barium chloride or calcium chloride, etc., can improve the reducing ability of NaBH4 and the selectivity of products, 9-fluorene methanol The selectivity was close to 100%, and the yield was up to 98.14%. The other components phenanthrene, anthracene and carbazole do not participate in the reaction of fluorene, and can be separated from fluorene during the reaction. Using fluorene and phenanthrene model compounds as raw materials, after reaction and separation, the purity of phenanthrene was 96.90%, and the yield was 98.95%; the purity of 9-fluorenemethanol was 97.68%, and the yield was up to 91.58%. Using anthracene slag as raw material, due to the accumulation of other components in the pretreatment process, the purities of phenanthrene and 9-fluorenemethanol obtained were 83.88% and 94.35%, and the yields were 64.46% and 89.49%, respectively.