experiment to study of formation of ester
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basically ... Easter is formed when a carboxylic acid reacts with alcohol.
experiment to study the formation of Easter - the reaction of ethanoic acid with ethanol in the presence of hot concentrated H2SO4 gives rise to the formation of Easter
CH3COOH + C2H5OH --------> CH3COOC2H5 (Easter) + H2O in the press of Δ concentered H2SO4
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experiment to study the formation of Easter - the reaction of ethanoic acid with ethanol in the presence of hot concentrated H2SO4 gives rise to the formation of Easter
CH3COOH + C2H5OH --------> CH3COOC2H5 (Easter) + H2O in the press of Δ concentered H2SO4
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Answered by
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Mix ethanol and ethanoic acid in a test-tube along with sulphuric acid.
Heat the mixture well.
CH3COOH + C2H5OH------->CH3COOC2H5 + H2O.
Here CH3COOC2H5 is ethyl ethanoate and is the ester.
This is esterification reaction.
The water produced is removed by sulphuric acid which acts as a dehydrating agent.
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Heat the mixture well.
CH3COOH + C2H5OH------->CH3COOC2H5 + H2O.
Here CH3COOC2H5 is ethyl ethanoate and is the ester.
This is esterification reaction.
The water produced is removed by sulphuric acid which acts as a dehydrating agent.
Feel free to message me if you have any doubts.
May this year be prosperous and glorious for you.
Please mark my answer as brainliest.
- Maheshwaran.
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