Edexcel iGCSE Chemistry -4.41–4.42C Structures of Esters , Properties and Uses of Esters- Study Notes- New Syllabus

4.41C Writing Structural and Displayed Formulae of Esters from Alcohols and Carboxylic Acids (and Vice Versa)

Esters are formed by the reaction between:

• An alcohol (contains −OH)
• A carboxylic acid (contains −COOH)

This reaction is called esterification.

\( \mathrm{Carboxylic\ acid + Alcohol \rightleftharpoons Ester + Water} \)

The Ester Functional Group

\( \mathrm{-C(=O)O-} \)

When forming an ester:

• The −OH from the acid is removed.
• The H from the alcohol’s −OH is removed.
• These combine to form water.
• The remaining parts join through an −O− link.

Step-by-Step Method (From Acid + Alcohol to Ester)

Example: Ethanoic acid + Propanol

Ethanoic acid:

\( \mathrm{CH_3COOH} \)

Propan-1-ol:

\( \mathrm{CH_3CH_2CH_2OH} \)

Remove:

• −OH from the acid
• H from the alcohol

Join remaining parts:

\( \mathrm{CH_3COOCH_2CH_2CH_3} \)

Name: propyl ethanoate

Naming Rule for Esters

• The alcohol gives the first part of the name (alkyl group).
• The acid gives the second part (ending changes to -oate).

Example:

• Ethanol → Ethyl
• Ethanoic acid → Ethanoate
• Product → Ethyl ethanoate

Working Backwards (From Ester to Acid + Alcohol)

Example: Ethyl propanoate

• “Ethyl” comes from ethanol
• “Propanoate” comes from propanoic acid

Alcohol:

\( \mathrm{C_2H_5OH} \)

Acid:

\( \mathrm{CH_3CH_2COOH} \)

Displayed Formula Example

Propyl ethanoate:

\( \mathrm{CH_3 – C(=O) – O – CH_2 – CH_2 – CH_3} \)

Must clearly show:

• The C=O double bond.
• The −O− linkage.
• All atoms.

 Key Points

• Alcohol part becomes the first name.
• Acid part changes to -oate.
• Remove −OH from acid and H from alcohol.
• Join through −O−.