The Aldol Condensation Reaction: Preparation of Benzalacetophenones

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AS202 Pure Chemistry

CHM556 Organic Chemistry II

Experiment 4

The Aldol Condensation Reaction:

Preparation of Benzalacetophenones (Chalcone).

Student’s name: ABDUL NIAT MODEN (2016644598)

Partner’s name: MOHD FARIS BIN ABDUL RASHID

Date: 16/11/2017

Submission date: 23/11/2017

Lecturer’s name: Madam Nur Vicky binti Bihud

OBJECTIVES:

Performing aldol condensation reaction between 3-nitrobenzaldehyde and acetophenone in the presence of base (ethanol) to produce α, β-unsaturated ketone (3-nitrochalcone) and calculate the percentage yield of the crude and purified product and determine the melting point and spectrum.

INTRODUCTION:

Chalcone is an aromatic ketone that forms the central core for a variety of important biological compounds. Other names for chalcone are benzalacetophenone and phenyl styryl ketone. Chalcones show antibacterial, antifungal, antitumor and anti-inflammatory properties. Chalcone can be prepared by an aldol condensation between a benzaldehyde and an acetophenone in the presence of a catalyst. Aldol condensation is also known as Claisen-Schmidt reaction. The aldol condensation relies on the reactivity of a carbonyl group to build a new carbon-carbon bond. The aldol reaction is one of the most powerful methods available for forming a carbon-carbon bond. In this reaction, the conjugate base of an aldehyde or ketone adds to the carbonyl group of another aldehyde or ketone to give a β-hydroxyaldehyde or β-hydroxyketone product. This is the intermediate product of the crossed-aldol reaction. A crossed-aldol condensation leads to a number of different products unless one of the carbonyl compounds involved cannot form an enolate ion which means the compound has no α-hydrogens. A good choice for such a compound is an aromatic aldehyde. This is because only one enolate ion will form, which is from other carbonyl compound. Once formed, the nucleophilic enolate ion attacks carbonyl carbon to form a β-hydroxycarbonyl product. The β-hydroxycarbonyl product then eliminates a molecule of water to form a conjugated system composed of a double bond and the carbonyl group.

Experiment Procedure:

1. 0.75g of 3-nitrobenzaldehyde was weighed and placed it into 50 ml Erlenmeyer flask.
2. 0.60ml of acetophenone and 4.0ml of 95% ethanol to the flask containing the aldehyde.
3. The flask been swirl to mix the reagents and dissolve the solids. The solution should be cooled to room temperature before proceeding to the next step.
4. 0.5ml of NaOH solution was added (prepared by the lab assistance using 6.0g of NaOH in 10ml of water.
5. The mixture was stirred until it solidifies or until the entire mixture becomes very cloudy.
6. 10ml of ice water added to the flask. If a solid is present at this point, break up the solid mass by stirring with a spatula. Cooling and scratching is necessary to initiate crystallization.
7. The mixture was transferred to a beaker with 15ml of ice water.
8. The solid was collected by vacuum filtration on a Buchner funnel. Wash with cold water and allowed it to air dry for 30 minutes.
9. The solid was weighed and determine the crude percent yield.
10. 0.5g of your solid was recrystallize from 10 ml of hot methanol. Cooling and scratching is necessary to initiate crystallization. The solid collected by vacuum filtration and dry in the oven.
11. The dry pure product was weighed and the percent yield was determined.
12. The melting point was measured and a proton NMR spectrum of the product was obtained.

Data:

Calculation:

Molar mass of acetophenone = 120 g/mol

Molar mass of 3-nitrobenzaldehyde = 151 g/mol

Molar mass of 3-nitrochalcone = 252g/mol

Density of acetophenone = 1.03 g/mol

Mol of 3-nitrobenzaldehyde = mass/ molar mass

                                = 0.7707/151

                                = 5.1026x10-3 mol

Mass of acetophenone = 0.6x 1.03