For the "Borohydride Reduction of Vanillin to Vanillyl Alcohol" reaction, the equation is:(NOT looking for the procedure, but the hypothetical amount of each compound needed)

 

4(vanillin) + NaBH4 +4H20 --(in NaOH solution)-->> 4(Vanillyl alcohol) + H3BO3 + 4NaOH

 

If the starting material is 25 mmol of vanillin, how much

Sodium Borohydride___,

Sodium Hydroxide___

HCl___

is needed to complete this reaction?

 

25mmolx(1 mol/1000mmol)x(152.2 M.W./1 mol)=.4g (400mg) of vanillin (the starting material). But from here, can someone show how much the rest of the chemicals are needed?

 

The Physical Properties

Vanillin - 152.2 M.W.

Sodium Borohydride - 37.83 M.W.

Vanillyl Alcohol -154.2 M.W.

(PLEASE show the calculations )

 

Oh, and for HCl it's (3 M), and for NaOH (1 M)

your gana need 3-4g sodium hydroxide as far as the HCL acid your gana wana either use a diluted solution of 1 to 3 or 4ml HCl