Hydrolysis of an ester

Recall 3 facts about triglycerides

Learning objectives Recall how to name triglycerides Describe how to undertake hydrolysis of an ester Explain the chemistry behind the reaction

How to name a fatty acid Count the carbon atoms Count the double bonds Say where the double bonds are (counting from the COOH group end) e.g. CH3 (CH2)12 COOH 14 C atoms so tetra (4) + Deca (10) = tetradecanoic acid, 14, 0 There are no double bonds e.g. CH3(CH2)5 CH=CH(CH2)7COOH 16 C atoms 1 double bond between the 9th and 10th C so hexa (6) + dec(10) = hexadec-9-enoic acid, 16, 1 (9)

Name the compounds CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH CH3CH2CH=CHCH2CH=CHCH2CH=CH(CH2)7COOH

Name the compounds CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH octadec-9,12-enoic acid, 18,2 (9, 12) CH3CH2CH=CHCH2CH=CHCH2CH=CH(CH2)7COOH octadec-9, 12, 15-enoic acid, 18, 3 (9, 12, 15)

Learning objectives Recall how to name triglycerides Describe how to undertake hydrolysis of an ester Explain the chemistry behind the reaction

Hydrolysis of an ester

Hydrolysis is the opposite of esterification HYDROLYSIS OF ESTERS Hydrolysis is the opposite of esterification ESTER + WATER CARBOXYLIC ACID + ALCOHOL HCOOH + C2H5OH METHANOIC ETHANOL ACID ETHYL METHANOATE CH3COOH + CH3OH ETHANOIC METHANOL ACID METHYL ETHANOATE

HYDROLYSIS OF ESTERS Hydrolysis is the opposite of esterification ESTER + WATER CARBOXYLIC ACID + ALCOHOL The products of hydrolysis depend on the conditions used... acidic CH3COOCH3 + H2O CH3COOH + CH3OH alkaline CH3COOCH3 + NaOH  CH3COO¯ Na+ + CH3OH If the hydrolysis takes place under alkaline conditions, the organic product is a water soluble ionic salt The carboxylic acid can be made by treating the salt with HCl CH3COO¯ Na+ + HCl  CH3COOH + NaCl

This is the equipment: What type of hydrolysis will you be doing List the basic techniques you will be doing and the order in which you will be doing them 3 cm3 methyl benzoate (harmful by inhalation) 10 cm3 2 mol dm–3 sodium hydroxide (corrosive) 10 cm3 ethanol (highly flammable) 20 cm3 2 mol dm–3 hydrochloric acid (irritant) Methyl orange indicator 50 cm3 round-bottomed flask (Quickfit) Water-cooled condenser (Quickfit) and rubber tubing Four 10 cm3 measuring cylinders Anti-bumping granules Bunsen burner and safety mat Stand, clamp and boss Two 100 cm3 beakers Stirring rod and spatula Access to vacuum filtration, a balance, melting point apparatus and boiling water Dropping pipettes

Thinking about the experiment. The solution is acidified after the reflux, why? Explain why ethanoic acid is water soluble but benzoic acid is not soluble in cold water. Suggest what might be the most significant procedural error and suggest a modification for it. Give two reasons why your percentage yield is less than 100%. A suggested mechanism for the hydrolysis of the ester involves the hydroxide ion as a nucleophile. Complete the mechanism below, putting dipoles and curly arrows where necessary.

Alkaline hydrolysis of an ester produces the carboxylate Alkaline hydrolysis of an ester produces the carboxylate. The solution was acidified to convert the carboxylate into the carboxylic acid. The polarity of the bonds in the COOH group in ethanoic acid produces interactions with polar water. As well as dipole-dipole interactions there will be hydrogen bonding between the acid and water. In benzoic acid the large benzene ring is non- polar. The delocalised π-electrons are attracted to the carbon in the COOH group, which reduces the polarity of the bonds. This means that the intermolecular interactions and the solubility of benzoic acid are also reduced. The volume of methyl benzoate is measured in a 10 cm3 measuring cylinder and as the yield is based on this measurement it would more appropriate to measure it in a graduated pipette. 4. The yield will be less than 100% because: The methyl benzoate may not be completely hydrolysed when it is refluxed. Some of the product is lost during each step of the purification process.