Q.Explain the following reactions (write only chemical equations):

Decarboxylation – The Big Picture

Imagine a carboxylic acid as a molecule with a long carbon chain that ends in a COOH group — a carbon double-bonded to one oxygen, single-bonded to an OH. That COOH is the "business end." Now picture heating that molecule so hard that the entire COOH group rips off, taking one extra carbon with it, and flies away as carbon dioxide ($CO_2$). What remains is a hydrocarbon with one fewer carbon than you started with.

That's decarboxylation: the loss of carbon dioxide from a carboxylic acid (or its salt).

The name itself tells the story: de- (remove) + carboxyl (the COOH group) + -ation (process). You are literally removing the carboxyl group as $CO_2$.

The Precise Statement

In words: When a carboxylic acid is heated strongly (often with a catalyst or base), the carboxyl group breaks off as carbon dioxide, and the hydrogen that was on the OH ends up attached to the carbon that was next to the COOH. The product is an alkane (or sometimes an alkene, depending on conditions).

But there's a catch: plain carboxylic acids don't decarboxylate easily. You need special conditions.

The Classic Exam Reaction: Soda-Lime Decarboxylation

The most common method you'll see in Indian exams is heating a sodium salt of a carboxylic acid with soda lime (a mixture of NaOH and CaO). The calcium oxide doesn't react — it just provides a porous, high-surface-area solid that helps the reaction go smoothly.

Here's the step-by-step:

1. Start with a carboxylic acid, say sodium acetate: $CH_3COONa$
2. Mix it with soda lime ($NaOH + CaO$) and heat strongly
3. The sodium salt reacts with NaOH to give methane and sodium carbonate:

The $CO_2$ that leaves is trapped as part of the sodium carbonate. The alkane (methane, in this case) escapes as a gas.

Why Does This Happen?

The carboxyl group is already partially "pre-organized" to lose $CO_2$. The carbon in $CO_2$ is in its highest oxidation state (+4), so it's very stable once formed. The driving force is the formation of that extremely stable $CO_2$ molecule plus a new C–H bond in the alkane.

In the soda-lime method, the NaOH provides the hydrogen that ends up on the alkane. The CaO helps by keeping the mixture dry and preventing side reactions.

A Common Mistake to Avoid …