Q.An aqueous solution of a substance on electrolysis gives ethane, then the substance is

Kolbe Electrolysis: From Intuition to Precision

Imagine you have a carboxylic acid — say, vinegar (acetic acid). You know it has a carboxyl group ($-COOH$) at one end. Now, what if you could snap off that carboxyl group and join the two remaining hydrocarbon pieces together? That is exactly what Kolbe electrolysis does: it takes two carboxylic acid molecules, removes their $CO_2$ groups, and couples the leftover alkyl fragments into a longer hydrocarbon chain.

The reaction happens in an electrolytic cell — the same kind of setup you use to split water into hydrogen and oxygen. But here, the "fuel" is a concentrated solution of a carboxylate salt (the conjugate base of the acid), and the electrodes are usually platinum.

The Core Idea in One Sentence

The carboxylate ions lose electrons at the anode, lose $CO_2$, and the two alkyl radicals ($R\cdot$) combine to form a dimer ($R-R$).

Step-by-Step Mechanism (Anode Only — That's Where the Action Is)

1. At the anode (oxidation): The carboxylate ion $RCOO^-$ gives up one electron to the electrode, forming a carboxyl radical:

2. Decarboxylation (loss of $CO_2$): The carboxyl radical is unstable. It immediately loses $CO_2$ to produce an alkyl radical:

3. Dimerization: Two alkyl radicals meet and couple:

The net result: two carboxylate ions become one alkane (the dimer) and two molecules of $CO_2$.

What You Actually See in the Lab

• Starting material: A concentrated aqueous or methanolic solution of the sodium or potassium salt of a carboxylic acid (e.g., sodium acetate, $CH_3COONa$).
• Electrodes: Inert platinum (carbon works too, but can get messy).
• Products at anode: The alkane dimer bubbles out (if short-chain) or deposits as a solid (if long-chain), along with $CO_2$ gas.
• Products at cathode: Hydrogen gas and hydroxide ions (the solution becomes basic).

For sodium acetate ($R = CH_3$), the product is ethane ($CH_3-CH_3$).

For sodium propionate ($R = CH_3CH_2$), the product is butane ($CH_3CH_2-CH_2CH_3$).

The Precise Statement (Exam-Ready)

Key Conditions and Limitations

• Concentration matters: The solution must be concentrated. In dilute solution, the carboxylate radical may instead react with water to form an alcohol or aldehyde (the Hofer–Moest reaction).
• No other oxidisable groups: If the alkyl chain has functional groups that are easier to oxidise (like $-OH$, $-NH_2$, or double bonds), those will react first — the reaction fails.
• Only symmetrical dimers: You get $R-R$ from $RCOO^-$. If you mix two different carboxylates ($RCOO^-$ and $R'COO^-$), you get a statistical mixture of $R-R$, $R-R'$, and $R'-R'$ — not useful for a single product. …