Difficulty: Intermediate | Prerequisites: Basic organic reaction mechanisms, stoichiometry, molarity calculations
This experiment synthesises tetraphenylcyclopentadienone (TPCPD) via a double aldol condensation of benzil and 1,3-diphenylacetone. Two sets of reaction conditions are compared side by side: one uses KOH in ethanol, the other uses Triton B in triethylene glycol. The point is to evaluate which method gives better yield, purity, and safety profile, and to decide which you would scale to 1 kg.
Benzil
A diketone (1,2-diphenylethane-1,2-dione) with molecular formula C₁₄H₁₀O₂ and a molecular weight of 210.23 g/mol. Think of it as two phenyl groups connected by two adjacent C=O groups. It supplies the two carbonyl carbons that end up in the five-membered ring.
1,3-Diphenylacetone (1,3-diphenyl-2-propanone)
A ketone (C₁₅H₁₄O, MW 210.27 g/mol) with two phenyl-bearing methylene groups flanking a central carbonyl. In simple terms, this is the molecule whose alpha-hydrogens get pulled off by base to form the enolate nucleophile.
Tetraphenylcyclopentadienone (TPCPD)
The product of the double aldol condensation, a deep-purple crystalline solid (C₂₉H₂₀O, MW 384.47 g/mol, lit. m.p. 218–220 °C). In simple terms, it is a five-membered ring bearing four phenyl groups and one ketone, formed by joining benzil and 1,3-diphenylacetone with loss of two equivalents of water.
Aldol condensation
A reaction in which an enolate (or enol) attacks a carbonyl carbon, forming a beta-hydroxy carbonyl intermediate that then loses water to give an alpha,beta-unsaturated carbonyl product. Think of it as "enolate attacks carbonyl, then dehydration."
KOH (potassium hydroxide)
A strong inorganic base used in Reaction A (2.0 M aqueous solution in ethanol as solvent). It deprotonates the alpha-hydrogens of 1,3-diphenylacetone to generate the enolate.
Triton B (benzyltrimethylammonium hydroxide)
A quaternary ammonium hydroxide used as the base in Reaction B (40 wt% solution in methanol, used with triethylene glycol as solvent). It is a bulky, soluble organic base. The advantage over KOH is that it dissolves more readily in the organic reaction medium and can give a more homogeneous reaction.
Theoretical yield
The maximum mass of product that could form if the limiting reagent were converted completely and no product were lost. Calculated from the moles of limiting reagent multiplied by the molar mass of the product.
Limiting reagent
The reactant present in the smallest stoichiometric amount, which determines the maximum possible yield. In this experiment, you compare moles of benzil and 1,3-diphenylacetone (1:1 ratio) to identify which one limits.
Atom economy
The ratio of the molecular weight of the desired product to the total molecular weight of all products (including by-products such as water). For TPCPD synthesis, atom economy is about 91% because the only by-product is water.
E-factor (Environmental factor)
Total mass of waste generated per mass of desired product. A lower E-factor means a greener process. Unlike atom economy, E-factor accounts for solvents, excess reagents, and all other waste.
Benzil and 1,3-diphenylacetone react in a 1:1 molar ratio in the presence of a base. The base deprotonates an alpha-hydrogen on 1,3-diphenylacetone, generating an enolate. That enolate attacks one of benzil's carbonyl carbons (nucleophilic addition), forming a beta-hydroxy intermediate that dehydrates. The sequence happens twice (once at each alpha-carbon of 1,3-diphenylacetone, once at each carbonyl of benzil), closing the five-membered ring and yielding TPCPD plus two equivalents of water.
The product is a distinctive deep-purple solid, which makes it easy to identify visually.
The experiment runs the same reaction under two different sets of conditions so students can compare outcomes.
Reaction A: KOH / Ethanol
Base: 2.0 M aqueous KOH (approximately 15 drops, or about 0.75 mL)
Solvent: ethanol
Heating: water bath
KOH is a strong, inexpensive inorganic base, but it has limited solubility in organic media, so the reaction mixture can be heterogeneous
Reaction B: Triton B / Triethylene Glycol
Base: 40 wt% Triton B (benzyltrimethylammonium hydroxide) in methanol
Solvent: triethylene glycol
Heating: sand bath (higher temperatures achievable)
Triton B is more soluble in organic solvents, giving a more homogeneous mixture. Triethylene glycol has a much higher boiling point than ethanol, allowing the reaction to run at elevated temperatures
The worksheet asks you to fill in a table for each reagent. Here is what you need and where to find it.
Molecular weights: benzil = 210.23 g/mol, 1,3-diphenylacetone = 210.27 g/mol, KOH = 56.11 g/mol, TPCPD = 384.47 g/mol
Grams: weigh your starting materials on the balance; for KOH solution, calculate from volume and molarity; for Triton B, calculate from volume and wt%
Mmols: grams divided by molecular weight, multiplied by 1000
Density and mL: relevant for liquid reagents (KOH solution, ethanol, triethylene glycol, Triton B solution). Solids (benzil, 1,3-diphenylacetone, TPCPD) do not require volume or density entries
Boiling and melting points: look these up from your textbook or SDS. The melting point of TPCPD (lit. 218–220 °C) is the key purity indicator
To make 20 mL of 2.0 M KOH:
Calculate mass needed: M × V = mol, so 2.0 mol/L × 0.020 L = 0.040 mol. Mass = 0.040 mol × 56.11 g/mol = 2.24 g
Dissolve 2.24 g of solid KOH in a small amount of distilled water, then dilute to 20 mL total volume in a volumetric flask
Dissolving KOH is highly exothermic, so add it slowly and use caution
Triton B is benzyltrimethylammonium hydroxide: a benzyl group (C₆H₅CH₂–) attached to a nitrogen bearing three methyl groups, with hydroxide as the counterion. Its main advantage over KOH is superior solubility in organic solvents, which makes the reaction mixture more homogeneous and can improve contact between base and substrate.
Compound | MW (g/mol) |
|---|---|
Benzil | 210.23 |
1,3-Diphenylacetone | 210.27 |
KOH | 56.11 |
TPCPD | 384.47 |
Water (by-product) | 18.02 |
15 drops of 2.0 M KOH is approximately 0.75 mL.
mmol = M x V(L) x 1000 = 2.0 x 0.00075 x 1000 = 1.5 mmol KOH
Triton B is supplied as a 40 wt% solution. To calculate mmol, you need the volume used, the density of the solution, and the MW of Triton B (167.25 g/mol for the hydroxide form).
Mass of Triton B = volume (mL) x density (g/mL) x 0.40
mmol = mass of Triton B / MW x 1000
The reaction is 1:1 (benzil : 1,3-diphenylacetone). Identify the limiting reagent by comparing mmol of each. Then:
Theoretical yield (g) = mmol of limiting reagent x MW of TPCPD / 1000
Theoretical yield (g) = mmol(limiting) x 384.47 / 1000
% yield = (actual yield / theoretical yield) x 100
Students often forget that 15 drops of KOH solution is not 15 mmol. You must use molarity and volume to get the correct mmol of base. The volume of 15 drops is approximately 0.75 mL, not 1 mL.
Students sometimes treat the KOH and Triton B as reagents consumed in the overall balanced equation. The base is a catalyst for enolate formation; it is not part of the stoichiometry of the product.
A high atom economy (91%) does not mean the process is green overall. Atom economy ignores solvents, excess reagents, and all non-product mass that must be disposed of. E-factor captures these.
Students mix up limiting reagent identification when the two starting materials have nearly identical molecular weights (210.23 vs 210.27). The grams may look similar, but always convert to moles and compare.
⚠️ You will likely be asked to compare the two reaction conditions (KOH/ethanol vs Triton B/triethylene glycol) on the basis of yield, purity, safety, waste, and scalability. Be ready to argue which you would scale to 1 kg and justify it.
⚠️ Atom economy vs E-factor is a recurring exam topic. Know that atom economy is a theoretical best-case metric that ignores solvents, while E-factor is an empirical measure of total waste per gram of product.
⚠️ Melting point range as a purity indicator: a narrow melting point range close to the literature value (218–220 °C) indicates high purity. A broad or depressed range suggests impurities.
⚠️ Preparing a solution of known molarity from a solid (e.g. making 2.0 M KOH from solid KOH) is a fundamental lab calculation that appears frequently on practical exams.
True or False: The by-product of the TPCPD synthesis is carbon dioxide. (False. It is water.)
Fill in the blank: 15 drops of 2.0 M KOH is approximately ______ mmol of KOH. (1.5 mmol)
True or False: Atom economy accounts for solvent waste. (False. It only considers the atoms in the balanced equation.)
Fill in the blank: The literature melting point of TPCPD is ______. (218–220 °C)
True or False: Triton B is preferred over KOH in some reactions because it is more soluble in organic solvents. (True.)
Q: Describe how you would prepare 20 mL of 2.0 M KOH from solid KOH.
A: Calculate the mass: 2.0 mol/L x 0.020 L = 0.040 mol, and 0.040 mol x 56.11 g/mol = 2.24 g. Dissolve 2.24 g of KOH pellets in a small volume of distilled water (the process is exothermic, so add slowly), then transfer to a 20 mL volumetric flask and dilute to the mark.
Q: Why is the atom economy of the TPCPD synthesis 91%, yet this figure does not tell the whole story?
A: Atom economy only considers the atoms in the balanced equation. The only by-product is water (2 x 18.02 g/mol = 36.04), so the ratio of product MW to total product MW is 384.47 / (384.47 + 36.04) = 91.4%. It ignores solvents, excess reagents, catalysts, and any waste from purification, all of which contribute to the real environmental impact. E-factor and PMI capture these.
Q: What is the structural advantage of Triton B over KOH as a base in this reaction?
A: Triton B (benzyltrimethylammonium hydroxide) is an organic quaternary ammonium base that dissolves readily in organic solvents. This gives a more homogeneous reaction mixture compared to the partly heterogeneous KOH/ethanol system, potentially improving contact between base and substrate.
Q: In the Discussion section, you are asked which method you would scale to 1 kg. What five criteria should your comparison address?
A: (a) Experimental procedure (ease, number of steps, heating method), (b) safety and hazards of the chemicals (consult SDS sheets), (c) waste generated (type and volume), (d) purity of product (melting point range), and (e) yield (percent yield of product).
Q: Calculate the E-factor for a reaction that produces 0.35 g of TPCPD and generates 4.2 g of total waste.
A: E-factor = total waste / mass of product = 4.2 / 0.35 = 12. This means 12 grams of waste are produced for every gram of product.
This experiment connects directly to aldol condensation mechanisms covered in the carbonyl chemistry unit. The enolate formation step is the same process you encounter in any base-catalysed aldol reaction, and the dehydration step parallels E1cb elimination.
The green chemistry metrics (atom economy, E-factor, PMI) tie back to the vanillin lab report and the broader theme of sustainable synthesis. Understanding why atom economy alone is insufficient prepares you for any exam question about evaluating the environmental impact of a reaction.
The comparison of reaction conditions (choice of base, solvent, heating method) is a transferable skill: industrial chemistry constantly evaluates trade-offs between cost, safety, yield, and waste when scaling up.
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