Question

At one point in the procedure, dilute hydrochloric acid is used to neutralize the solution. Specifically, what is being neutralized? (there is more than one answer here.)

Procedure:

Place 200 mg sodium methoxide and 400mg sodium borohydride into a 50mL beaker with 10mL Methanol. Stir the contents of the beaker genetly and let it sit undistrurbeded for 15minutes.

Preparation of the 9-fluorenone solution

Into a 250mL beaker place a 500mg 9-fluorenone, crushing any large pices of the solid. Add 10mL 95% ethanol and stir to dissolve the most of the 9-fluoronone. This solution will be yellow in color. The progress of the reaction can be followed by observing the color change since the product, 90-fluorenol, is white and will be colorless in solution.

Add the reducing solution dropwise (1drop /5sec) to the ketone solution while swirling gently. After addition is complete, and the color change has occured (it may take up to 30minutes), slowly add 20mL water, then neutralize the resulting solution with 3.0M HCl. Filter the resulting solid using a buchner funnel and wash the crystals using a few milliliters of cold DI water. Set the crystals out to dry on filter paper under the heat lamp. Weigh the final product. Determine its melting point range and obtain an infrared spectrum.

Answer 1

Hi, you are adding water and HCl to quench the reaction by neutralizing it and "destroying" any borohydride left in the solution. I checked with the masses and found that you have 2.78 mol of 9-fluorenone while you have 10.5 mol of sodium borohydride which means that your reducing agent is in excess (actually four times that of reagent to ensure complete reduction). So your HCl nuetralized the sodium methoxide

Your 9-fluorenone is not directly reduced to the alcohol, instead it is more like an alkoxide which should be reduced. So yes, you are nuetralizing your product alkoxide to alcohol. Note that it might not look like an alkoxide to you since the negative charge is on boron, but oxygen has the higher electronegativity and it has a negative charge so it accepts a proton to expel the BH3.

And finally any remaining NaBH4 is destroyed using HCl