Is vanillin soluble in NaOH?

Is vanillin soluble in NaOH?

Is vanillin soluble in NaOH?

Vanillin is mostly insoluble in water but soluble in aqueous sodium hydroxide solution? What did you add to make the vanillin dissolve and explain why it dissolves. What four purposes did aqueous hydrochloric acid serve in the reduction of vanillin to vanillyl alcohol?

Why is NaOH used in the reduction of vanillin?

Since sodium hydroxide is the conjugate base of water (pKa = 15.7), the deprotonation of vanillin will be essentially quantitative, giving a homogenous solution. In theory, one equivalent of NaBH4 could reduce four equivalents of vanillin since all four of the hydrides could react.

Why should the sodium borohydride be added to the vanillin in dropwise fashion?

Why is it important to add the sodium borohydride solution dropwise? It is important to add the sodium borohydride solution dropwise because too much of the solution that is added in a short period of time, will create a lot of heat. Sodium borohydride reacts violently as well, when added to the mixture too quickly.

Why is vanillin soluble in water?

Molecules containing only carbon and hydrogen are mostly insoluble in water. The oxygen-containing groups attached to the ring in vanillin can form strong hydrogen bonds with water, making it water soluble (about a gram of vanillin can be dissolved in 100 mL of cold water).

Why is vanillin soluble?

The oxygen-containing groups attached to the ring in vanillin can form strong hydrogen bonds with water, making it water soluble (about a gram of vanillin can be dissolved in 100 mL of cold water).

Is vanillin the limiting reagent?

Vanillin is the limiting reagent in the reactions conducted in this investigation, full conversion of the reaction is therefore reached when all vanillin has reacted.

What effect would doubling the concentration of sodium borohydride have on the rate of this reaction?

What effect would doubling the concentration of sodium borohydride have on the rate of this reaction? The rate would be doubled since this is a first‑order reaction.