Q.(a) How is acetic acid prepared from methyl magnesium bromide? What happens when it reacts with

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. …