Q.Sodium salt of which acid will be needed for the preparation of propane ? Write chemical equation for the reaction.

Concept and Intuition

In Kolbe's electrolysis, the carboxylate salt of a carboxylic acid is electrolyzed. At the anode, the carboxylate ion loses an electron to form a carboxyl radical, which instantly loses CO$_2$ to give an alkyl free radical; two such radicals combine (radical dimerization) to give a symmetric alkane with twice the carbon count of the alkyl group (not the acid). At the cathode, water is reduced, liberating H$_2$ gas.

Step-by-Step Solution

1. Sodium propionate is $\mathrm{CH_3CH_2COONa}$ — the propionate ion is $\mathrm{CH_3CH_2COO^-}$, so the alkyl group $R = \mathrm{C_2H_5}$ (ethyl, 2 carbons).
2. Kolbe coupling gives $R-R = \mathrm{C_2H_5-C_2H_5} = \mathrm{C_4H_{10}}$, i.e. n-butane, not n-hexane (hexane would require a 3-carbon alkyl group, i.e. from a butanoate/butyrate salt).
3. So Statement-I (claiming it gives n-hexane) is FALSE. …

Concept and Intuition

In Kolbe's electrolytic method, the carboxylate anion is oxidised at the anode: it loses an electron to form a carboxylate radical, which instantly loses $\mathrm{CO_2}$ to form an alkyl radical; two such alkyl radicals combine to give the alkane. Since two acetate ions are needed to make one ethane, the moles of $\mathrm{CO_2}$ produced are always twice the moles of alkane.

Step-by-Step Solution

1. Anode reaction: $2\mathrm{CH_3COO^-} \to 2\mathrm{CH_3COO}^{\bullet} + 2e^-$.
2. Each $\mathrm{CH_3COO}^{\bullet}$ radical decarboxylates: $\mathrm{CH_3COO}^{\bullet} \to \mathrm{CH_3}^{\bullet} + \mathrm{CO_2}$.
3. Two methyl radicals combine: $2\mathrm{CH_3}^{\bullet} \to \mathrm{CH_3-CH_3}$ (ethane).
4. Net: $2\mathrm{CH_3COO^-} \to \mathrm{C_2H_6} + 2\mathrm{CO_2} + 2e^-$. …

Concept and Intuition

Kolbe's electrolysis is an anodic oxidative decarboxylation: a carboxylate ion loses an electron at the anode, releases $CO_2$, and the resulting alkyl free radical dimerizes to give a symmetrical alkane with twice the carbon count of the alkyl group. Simultaneously, at the cathode, water (or $H^+$) is reduced, liberating hydrogen gas — this is the standard cathodic half-reaction in aqueous electrolysis.

Step-by-Step Solution

1. Sodium propanoate: $CH_3CH_2COO^-Na^+$; the anion is $CH_3CH_2COO^-$ (propanoate, 3 carbons including the carboxyl carbon).
2. At the anode: oxidation removes an electron, decarboxylation follows: $CH_3CH_2COO^- \rightarrow CH_3CH_2^\bullet + CO_2 + e^-$. This gives the ethyl radical ($C_2H_5^\bullet$).
3. Two ethyl radicals couple: $2\,C_2H_5^\bullet \rightarrow C_4H_{10}$ (butane) — this is the anode product. …

Concept and Intuition

Kolbe electrolysis is an oxidative radical-coupling reaction: carboxylate anions ($\mathrm{RCOO^-}$) migrate to the anode, get oxidized (lose an electron each) to form unstable carboxyl radicals $\mathrm{RCOO^\bullet}$, which instantly lose $\mathrm{CO_2}$ to give alkyl radicals $\mathrm{R^\bullet}$; two such radicals combine (couple) to form the symmetrical alkane $\mathrm{R\text{-}R}$. Meanwhile at the cathode, reduction of water (or $\mathrm{H^+}$) liberates hydrogen gas.

Step-by-Step Solution

1. Sodium propanoate: $\mathrm{CH_3CH_2COO^-Na^+}$, so R = $\mathrm{CH_3CH_2}-$ (ethyl group, 2 carbons).
2. At the anode: $2\,\mathrm{CH_3CH_2COO^-} \rightarrow 2\,\mathrm{CH_3CH_2COO^\bullet} + 2e^-$; each radical loses $\mathrm{CO_2}$ to give $\mathrm{CH_3CH_2^\bullet}$ (ethyl radical).
3. Two ethyl radicals couple: $2\,\mathrm{CH_3CH_2^\bullet} \rightarrow \mathrm{CH_3CH_2CH_2CH_3}$ (butane, X), a 4-carbon alkane. …

Concept and Intuition

Kolbe electrolysis is the anodic oxidation of carboxylate ions: at the anode, $RCOO^-$ loses an electron to form a carboxyl radical $RCOO^\bullet$, which instantly decarboxylates (loses $CO_2$) to give an alkyl radical $R^\bullet$; two such radicals combine (dimerize) to form the symmetric alkane $R-R$. This effectively couples two alkyl fragments while ejecting the original carboxyl carbons as $CO_2$.

Step-by-Step Solution

1. Sodium butanoate is $CH_3CH_2CH_2COO^-Na^+$ (4 carbons total, since butanoic acid is $C_3H_7COOH$).
2. At the anode: $2\,CH_3CH_2CH_2COO^- \rightarrow 2CO_2 + 2\,CH_3CH_2CH_2^\bullet$ (propyl radical, 3 carbons each) $+ 2e^-$.
3. The two propyl radicals combine: $CH_3CH_2CH_2^\bullet + {}^\bullet CH_2CH_2CH_3 \rightarrow CH_3CH_2CH_2-CH_2CH_2CH_3$ (n-hexane, $C_6H_{14}$). …

Concept and Intuition

In Kolbe's electrolysis, the carboxylate anion ($CH_3COO^-$) migrates to the anode (the positive electrode) where oxidation occurs. It loses an electron to become a carboxyl radical $CH_3COO^\bullet$, which instantly loses $CO_2$ to form a methyl free radical $CH_3^\bullet$; two such radicals combine (dimerize) to give ethane. Oxidation (loss of electrons from an anion) always happens at the anode, since the cathode is where reduction (gain of electrons, e.g. discharge of $H^+$ to $H_2$) occurs.

Step-by-Step Solution

1. Sodium acetate solution is electrolyzed; $CH_3COO^-$ ions migrate to the anode.
2. At the anode: $CH_3COO^- \to CH_3COO^\bullet + e^-$ (oxidation/loss of electron).
3. $CH_3COO^\bullet$ decarboxylates rapidly: $CH_3COO^\bullet \to CH_3^\bullet + CO_2$.
4. Two methyl radicals combine: $2CH_3^\bullet \to C_2H_6$ (ethane) — confirming the Assertion. …

Concept and Intuition

This question combines stoichiometric neutralization with the Kolbe electrolytic decarboxylation reaction. When a solution of a sodium carboxylate (here sodium acetate, from neutralizing acetic acid with NaOH) is electrolyzed, the carboxylate anion is oxidized at the anode: two carboxylate ions lose two electrons and two $CO_2$ molecules are released, while the two resulting alkyl radicals combine to form a symmetrical alkane. For acetate specifically:

So every 2 moles of acetate ion produce exactly 1 mole of ethane gas at the anode (plus $H_2$ at the cathode and $CO_2$, which are not asked about here).

Step-by-Step Solution

1. Calculate moles of acetic acid: $n = M \times V = 0.2\ mol/L \times 0.100\ L = 0.02\ mol$.
2. Complete neutralization with NaOH converts all the acetic acid into sodium acetate: moles of $CH_3COO^-$ = 0.02 mol.
3. Apply the Kolbe electrolysis stoichiometry: 2 mol $CH_3COO^-$ → 1 mol $C_2H_6$ (ethane) + 2 mol $CO_2$. …