Introduction
In the present post, I will describe the aldol condensation reaction between vanillin (4-hydroxy-3-methoxy benzaldehyde) and acetone (2-propanone) to yield the product (E)-4-(4-hydroxy-3-methoxyphenyl)but-3-en-2-one. The reaction is quite simple and involves commonly available items and reagents and represents a great way to demonstrate this type of reaction. The overall reaction can be summarized as follows:
The single substitution is ensured by operating in a large excess of acetone.
Reaction mechanism
The reaction mechanism is very simple and starts with the deprotonation of acetone to produce, in equilibrium, small quantities of the corresponding enolate ion:
The obtained enolate ion readily reacts with the more reactive, and non enolizable, benzaldehyde function of vanilin leading to an intermediate aldol addiction product carrying an alcoholic function on the $\beta$ carbon atom:
The obtained aldol addiction product can be further deprotonated on the $\alpha$ carbon leading to an enolate ion structure that can eliminate the hydroxy group by an E1cb mechanism. The elimination of water leads to the formation of a stable conjugated $\pi$ system that drives all the equilibrium forwards.
Experimental part
In an Erlenmeyer flask, 3.90g of vanillin were dissolved in 30mL of acetone. To the solution, 1.97g of sodium hydroxide, dissolved in 20mL of water, were added. Upon the addition of the base the solution turned bright yellow and kept darkening with the progress of the reaction. The mixture was left reacting under constant stirring for almost 24h. During this time the solution assumed a dark orange/red tint with bright yellow reflexes when swirled.
To the reaction mixture, 100mL of a 16% hydrochloric acid solution were added. Upon adding the acid, the solution heated up slightly and the color turned initially light yellow and then gradually switched to a red/brown color. At the end of the addition, the pH is strongly acidic but no sign of precipitate was observed.
The solution was left to cool down and the walls of the flask were scratched with a glass rod. A fine precipitate of greenish almost olive color is formed and recovered by gravity filtration. The obtained precipitate was washed with water and dried. The obtained solid was recrystallized from a hot 1:1 ethanol-water mixture leading, upon slow cooling, to a bright yellow crystalline precipitate.
As a side note, the presented procedure is not perfect and the most critical part of it is represented by the crystallization phase. Once the reaction mixture is acidified it is not unfrequent to observe the product crushing out of solution as a dark almost black oil instead of nice yellow crystals. Often an in between behavior is observed where a semi-crystalline product is obtained in which the yellow crystals are mixed with an underlying amorphous black/brown phase leading to precipitates with colors varying from yellow to brown to almost greenish color. The addition of the acid in diluted form seemed to help the crystallization process that however is still not perfect. Often adjusting the pH, boiling the solution to eliminate excess acetone, or increasing the ionic strength of the solution seemed to help the direct precipitation of the crystalline form or, sometimes, even the direct conversion of the black oil to a crystalline yellow solid. More experimentation is needed however to obtain a reliable procedure.
1H-NMR spectrum
The 1H-NMR spectrum has been recorded using a 200MHz instrument dissolving the sample in chloroform-d. The spectrum is consistent with the expected molecule. The singlet at 2.57ppm can be assigned to the methyl group at the $\alpha$ position to the carbonyl group while the other singlet at 4.13ppm can be assigned to the protons of the methoxy group directly connected to the ring. The broad singlet at 6.14ppm clearly belongs to the alcholic proton of the phenol. The trans protons located on the two sides of the double bonds are responsible for the two doublets of peaks at 6.78 and 7.65ppm splitted with a large coupling constant of 16.2Hz. The higher shift is clearly associated with the proton in $\beta$ position due to the resonance coupling of the electronic structure of the double bond to the carbonyl system. The other aromatic protons fall, together with the residual signal of chloroform, around 7.26ppm. Individually assigning these protons is not trivial from this spectrum.
FTIR spectrum
The Infrared spectrum of the molecule has been collected using the KBr disk method. The spectrum presents a broad peak at 3298 $cm^{-1}$ associated with the stretching of the phenol hydroxy group. The peaks in the region around 3000 $cm^{-1}$ are associated with the stretching of CH bonds of the molecule. The intense peaks at 1637, 1583, and 1519 $cm^{-1}$ are probably associated with the stretching of the C=O carbonyl bond and the C=C double bonds of the alkene and the aromatic ring. The extensive conjugation of the system probably accounts for the shift at lower energies of the vibrational transitions associated with such groups. The transitions at lower energies are difficult to assign.
