Hydrolysis of an ester

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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
1. Count the carbon atoms
2. Count the double bonds
3. 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
1. CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH
2. CH3CH2CH=CHCH2CH=CHCH2CH=CH(CH2)7COOH
Name the compounds
1. CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH
octadec-9,12-enoic acid, 18,2 (9, 12)
2. 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 OF ESTERS
Hydrolysis is the opposite of esterification
ESTER + WATER
CARBOXYLIC ACID + ALCOHOL
HCOOH
+
METHANOIC
ACID
C2H5OH
ETHANOL
ETHYL METHANOATE
CH3COOH
ETHANOIC
ACID
METHYL ETHANOATE
+
CH3OH
METHANOL
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:
1. What type of hydrolysis will you
be doing
2. 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.
1. The solution is acidified after the reflux, why?
2. Explain why ethanoic acid is water soluble but benzoic acid is not soluble
in cold water.
3. Suggest what might be the most significant procedural error and suggest a
modification for it.
4. Give two reasons why your percentage yield is less than 100%.
5. 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.
1. Alkaline hydrolysis of an ester produces the carboxylate. The solution was
acidified to convert the carboxylate into the carboxylic acid.
2. 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.
3. 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.

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