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Aldol Synthesis of Dibenzalacetone Essay

The goal of this experiment was to synthesize dibenzalacetone by aldol synthesis. The name ‘Aldol synthesis’ was taken from the words ‘aldehyde and alcohol’. This is because the product of this reaction contains both an aldehyde and alcohol. The carbon-carbon bond-forming reaction is referred to as aldol addition. An aldol condensation yields many species of products if the reactant is more than one. Therefore, the aldehyde has to react with itself to yield one product.

Procedure: Followed according to the lab manual.

Results:DataValueWeight of crude crystal1.6 gWeight of recrystallization crystal0.48 gColor of crystalOrange-yellow% recovery of recrystallization:(0.48 g)/(1.6 g) ×100 %=30%Theoretical yield (from prelab): 1.177 gPercent yield(0.48 g)/(1.177 g) ×100 %=40.78%


Through this experiment, aldehyde reacts with itself in the presence of acetone and base, NaOH. Acetone is used as the enolate forming compound, adding to the benzaldehyde followed by the dehydration to form a benzal group. As for the low yield (40.78%), it can be attributed to a few factors. The first being that this reaction goes by an equilibrium and that if this was not shifted far enough to the dibenzalacetone side, the reaction would not go to completion and less yield can be expected. It might be caused by the solvent used during recrystallization was a little bit more than advised in the lab manual (~15 mL instead of ~12 mL of ethanol). The crystal, (I’m afraid if) it would be solved beyond recrystallizing at this point; but it recrystallized as it cools down, so I was sure that the solvent used was not too much. The crystal was also wasted in the Erlenmeyer flask and some on the filter paper used to dry it.

IR analysisThe IR result from this experiment shows no significant stretches infrequency higher than 2000 cm-1 except for a trough at 3052.68 cm-1 that indicates an aromatic ring bond to hydrogen (Ar-H). Most of the expected stretches in this IR are in between 1652.41 cm-1 to 1595.95 cm-1. In this range, the stretches are most likely indicating the presence of alkenes, C=C and carbonyl group, C=O (or ketone).

H NMR analysisFrom the H NMR analysis the significant (and relevant) details shall be discussed. There are two doublets consisting of two hydrogen at 7.12 ppm (2848 MHz) and 7.08 ppm (2832 MHz). These most upfield protons as told in the prelab, is the least deshielded protons in the molecule. It is a doublets because there is a vicinal carbon with a proton, so it makes sense, according to the n + 1 rule. The next peaks about to be discussed are the multiplets between 7.32 ppm (2928 MHz) and 7.44 ppm (2976 MHz), consisting of 10 hydrogen. These protons belong to the two aromatic rings in the molecule, with five protons bonded to each ring.

The mysterious multiplets present could be interpreted also as the protons bonded to the phenyl group. But to make it relevant, the previous multiplets has to be reduced to a smaller range (see the H NMR graph, the first few peaks are not significant, so it could be ignored) of 7.41 ppm (2964 MHz) to 7.44 ppm (2976 MHz), and the other aromatic ring protons has a range of 7.61 ppm(3044 MHz) to 7.65 ppm (3060 MHz).

This explanation should be relevant to consider the multiplets are divided to 5H each, and to consider that these two benzene rings gives different signals as they conjugate, so the readings shifted a bit from each other. The last peaks are a doublets of two protons, which is the most downfield proton; the α hydrogen is indeed the most deshielded proton, as it is next to a carbonyl group – a withdrawing group. The doublets peak mirrors a vicinal carbon with one hydrogen; the hydrogen mention earlier in the analysis. Peaks are at 7.74 ppm (3096 MHz) to 7.78 ppm (3112 MHz).

To determine the double bond geometry, the chemical shift difference between two peaks in the doublets is taken to calculation. The difference of the first doublets is 0.04 ppm = 16 MHz. From the reference in the lab manual, this is a trans-configuration. The next doublets also has the same number of 3J value, which is 16 MHz – this indicates that the geometry of the dibenzalacetone in Trans (or E configuration) in both double bonds.

QuestionsNaOH catalyst and benzaldehyde should be added first, then the acetone. If the acetone goes in first, it could do aldol condensation on itself, in which enolate anions (of acetone) just attack neutral acetone carbonyls. This would yield a different product.


Organic Chemistry 7th EdFrancis A. CareyMcGraw Hill Publication

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