Question

Cooh -Ch2-Ch2-Cooh electrolysis

Answer 1

The main organic product of the electrolysis of succinic acid is Ethene ($CH_2=CH_2$). Other products include Carbon dioxide ($CO_2$), Hydrogen gas ($H_2$), and Sodium hydroxide ($NaOH$).

Answer 2

The electrolysis of succinic acid (COOH-CH2-CH2-COOH), specifically its disodium salt, is an example of Kolbe's electrolytic synthesis.

Mechanism:

1. Dissociation: In an aqueous solution, disodium succinate ($NaOOC-CH_2-CH_2-COONa$) dissociates into $2Na^+$ ions and succinate dianions ($^-OOC-CH_2-CH_2-COO^-$).

2. At the Anode (Oxidation):

   The succinate dianion migrates to the anode and loses two electrons, forming an acyloxy diradical.

   $^-OOC-CH_2-CH_2-COO^- \xrightarrow{-2e^-} \bullet OOC-CH_2-CH_2-COO^\bullet$

   This acyloxy diradical is unstable and undergoes decarboxylation, releasing two molecules of carbon dioxide ($CO_2$) and forming an ethylene diradical.

   $\bullet OOC-CH_2-CH_2-COO^\bullet \rightarrow \bullet CH_2-CH_2^\bullet + 2CO_2 \uparrow$

   The ethylene diradical immediately undergoes intramolecular coupling to form a double bond, yielding ethene ($CH_2=CH_2$).

   $\bullet CH_2-CH_2^\bullet \rightarrow CH_2=CH_2 \uparrow$

3. At the Cathode (Reduction):

   Water molecules are reduced, producing hydrogen gas ($H_2$) and hydroxide ions ($OH^-$).

   $2H_2O + 2e^- \rightarrow H_2 \uparrow + 2OH^-$

Overall Reaction (using the disodium salt):

$NaOOC-CH_2-CH_2-COONa(aq) + 2H_2O(l) \xrightarrow{electrolysis} CH_2=CH_2(g) + 2CO_2(g) + H_2(g) + 2NaOH(aq)$

Products:

• Anode: Ethene ($CH_2=CH_2$) and Carbon dioxide ($CO_2$)
• Cathode: Hydrogen gas ($H_2$) and Sodium hydroxide ($NaOH$) (in solution)

SMILES representation of involved structures:

• Succinic acid: O=C(O)CC(=O)O
• Ethene: C=C