Question

In a Sodium Borohydride reduction of 9-fluorenone to 9-fluorenol, methanol is used as the solvent because it will not react with the products and reactants and because the products and reactants are very sluble in methanol. Would 2-propanone also be a suitable solvent for this reaction? Why/Why not?

Answer 1

methanol=polar protic solvent( can form hydrogen bonds,have acidic hydrogen,can dissolve ionic compounds)

acetone(2-propanone)=polar aprotic solvent(cannot form hydrogen bonds,do not have acidic hydrogen,cannot dissolve ionic compounds

1)No, 2-propanone would not be a suitable solvent for this reaction,as the reaction involves the use of sodium borohydride(NaBH4) as the reducing agent to reduce the ketone functional group in 9-fluorenone to -OH group in the product 9-fluorenol.As 2-propanone itself is a ketone so it will react with the reducing agent and thus interfere with the reaction.

2) NaBH4 reducing agent is ionic in nature so needs a prolar protic solvent to stabilize it in solution ,by electrostatic interactions.Also acid is added for the work up of the final product, soluble in protic solvents and not aprotic solvents like H2SO4

Moreover , the solubility of 9-fluorenone and 9-fluorenol is very high in methanol , a polar protic solvent.It can form hydrogen bonding with the -OH and polar -C=O groups in reactant and product.

2-propanone (or acetone) is a polar aprotic solvent ,which can have dipole-dipole interactions with the reactant and product ,but the interactions are weaker compared to hydrogen bonding.So not good solvent.

check the reaction