Question

For Reduction of Vanillin aldehyde to Vanillyl alcohol using Nabh4, Aqueous NaOH and HCl

If you incorrectly add acid and do not attain a pH of 1, your yield may be lower than it should be as some of your product would not be isolated. Why won’t it be isolated?

Notes:

This addition of HCl causes four reactions to occur: a. it hydrolyzes the O-B bond in the intermediate and protonates the oxygen atom that was originally the oxygen atom in the carbonyl group b. it destroys any excess NaBH4 that may be present in the reaction mixture c. it neutralizes excess NaOH and d. it protonates the phenolic oxygen.

the reaction mixture is acidified with excess HCl, which causes the four reactions described previously to occur. Mention is made in Step 7 of the Experimental Procedure that vigorous bubbles of hydrogen gas will be observed when HCl is added, as a result of destroying excess NaBH4. The unbalanced equation for the reaction that occurs is H:! (from NaBH4) + H⊕ (from HCl) à H2 (gas). Because this is an acid-base reaction, heat is liberated and the reaction mixture becomes slightly warm. All boron atoms from NaBH4 are converted to boric acid, H3BO3, which is watersoluble. This and all other inorganic ions, such as Na⊕, H⊕, Cl!, are removed from the product by washing the solid product with water after it is collected by vacuum filtration. Washing with water also removes unreacted vanillin (if any) which is more soluble in water than the product. Since Vanillyl alcohol is not soluble in water, it will begin to precipitate from solution during the slow acidification. The process of crystallization is completed by cooling in an ice-water bath. The solid that crystallizes (vanillyl alcohol) is collected by vacuum filtration, washed with water and allowed to air dry.

Answer 1

Solution :-

If the acid is added incorrectly and pH of the solution is not attained to level pH= 1

Then it means there is no excess HCl which is required for the reactions to complete in each step

So if the HCl is not enough for the completion of the reactions then some of the intermediate will remain unconverted to product and remain dissolved in the water (solvent). Also if the NaOH is not neutralized because of the less HCl then NaOH will deprotonate the phenolic OH and forms soluble phenolic anion which will remain dissolved in the solvent therefore when the HCl added is not enough to complete all the reactions then some of the product remain in the form of intermediate and does not precipitate out so it get lost in the solvent because of this all the product is not isolated.

Following image shows the reactions of HCl in the reaction scheme.