Q.Write the name of the alcohol compound used in the preparation of the following ester compound: CH3-C(=O)-O-CH(CH3)-CH3 (i.e. CH3-CO-O-CH(CH3)2).

You already know that an acid and an alcohol can react. But why would they? And what do they make?

Think of a carboxylic acid as a molecule with a "sticky" end — the carboxyl group ($-COOH$). That end has an -OH group of its own. An alcohol also has an -OH group. If you bring them together with a little help from a strong acid catalyst (like concentrated $H_2SO_4$), the -OH from the acid and the -H from the alcohol's -OH group can break off and join to form water. The two remaining fragments snap together, forming a new molecule: an ester.

So the reaction is a kind of condensation — two molecules join by losing a small molecule (water). The catalyst (acid) makes the acid's carbonyl carbon more reactive so the alcohol can attack it.

Precise statement:

Esterification is the reversible, acid-catalyzed reaction between a carboxylic acid and an alcohol to produce an ester and water.

The general equation is:

Here, $R$ and $R'$ are alkyl (or aryl) groups. The $R$ comes from the acid, the $R'$ from the alcohol. The ester link is $-COO-$.

Key exam points:

• The reaction is reversible — the ester can be hydrolysed back to acid and alcohol (the reverse is called saponification when done with base).
• Concentrated $H_2SO_4$ acts as both a catalyst and a dehydrating agent (it absorbs the water formed, shifting equilibrium toward products).
• The reaction is slow at room temperature; heat is needed.
• Esters have characteristic fruity smells — that's why this reaction is used in perfumes and flavourings. …