Question

Product is 5-(4-hydroxy-3-methoxybenzal)creatinine. Will rate ASAP. Thank you!!

Questions: 1. Give the step-to-step mechanism for the condensation of vanillin and creatinine. (based on what you did for this experiment, decide if you have an acid or base catalyzed reaction) 2. This aldol condensation does not require a base. Why? 3. What is the major difference between the IR spectra of vanillin and the condensation product?

Answer 1

1. It is a type of the Aldol condensation which is very known in the literature that the driving force of this type of condensation is possible by three conditions as below,

a) Thermal condensation- only heating he two reactants ( vanillin and creatinine) neat at high temperature (155-170 ⁰C), the reaction proceeds by enolate formation of creatinine and react with vanillin. The formation of the water molecule confims the dehydation by elimination mechanism.

b) Base catalyzed condensation

c) Acid catalyzed condensation

The step to step possible base and acid catalyzed mechanism of the condensation of vanillin and creatinine is described in the details as below images.

Page 1 Condensaties of vanillis waith creatinine W) Thermal condensation ,155 24 V cho TUM И ON - И,0 Супе Dehydration product b) Base Catalysed to си Оме, stepe m enolate I step. 2 1 L Dehydration ome 5-(4-hyasury-3-methoxybenzal) creatinine product

Date / 102 bilt on 7 10 Sad Note Page c) Aeid Catalysed Condensation OY V Fome GY Enolate mo step 2 I stinomation Me an product

2. This aldol condensation do not require base because the above reaction can be possible simply by heating at high temperature. The starting material creatinine itself is basic in nature due to substituted guanidine as part structure which may responsible for enolate formation and dehydration process at   high temperature (155-170 ⁰C)

3. The major difference in the IR of reactant vanillin and product 5-(4-hydroxy-3-methoxybenzal)creatinine is that the characteristic peak at 1700 cm -1 in IR due to the aromatic aldehyde from vanillin was disappeared in the 5-(4-hydroxy-3-methoxybenzal)creatinine product. The additional peak at 1680-1620 cm-1 due to newly formed C=C as alpha beta unsaturation (alkene)