Question

Which of the following will not produce ethane? (1) reduction of \(\mathrm{CH}_{3} \mathrm{COOH}\) with \(\mathrm{HI}\) and red \(\mathrm{P}\) (2) reduction of \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) with HI and red \(\mathrm{P}\) (3) soda lime decarboxylation of sodium propionate (4) hydrogenation of ethene in the presence of Raney nickel

Short answer

Reduction of \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) with HI and red P will not produce ethane.

Step-by-step solution

Understand the relevant chemical reactions

Review the reactions and understand what products they yield. Compare each reaction to check which produces ethane.

Reduction of \(\mathrm{CH}_{3} \mathrm{COOH}\) with HI and red P

When acetic acid \(\mathrm{CH}_{3} \mathrm{COOH}\) is reduced with \mathrm{HI}\text{ and red phosphorus, it forms ethane } \mathrm{CH}_{3} \mathrm{CH}_{3}\.

Reduction of \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) with HI and red P

When acetone \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) is reduced with \mathrm{HI}\text{ and red phosphorus, it does not form ethane. It produces propane }\(\mathrm{CH}_3 \mathrm{CH}_{2} \mathrm{CH}_3\).

Soda lime decarboxylation of sodium propionate

Soda lime decarboxylation of sodium propionate forms ethane. The reaction is: \[\mathrm{CH}_{3} \mathrm{CH}_{2}\mathrm{COONa} + NaOH \xrightarrow{\text{soda lime}} \mathrm{CH}_{3} \mathrm{CH}_{3} + Na_2CO_3\]

Hydrogenation of ethene

Hydrogenation of ethene in the presence of Raney nickel produces ethane. The reaction is: \[ \mathrm{CH}_{2} = \mathrm{CH}_{2} + H_{2} \xrightarrow{\text{Raney Ni}} \mathrm{CH}_{3} \mathrm{CH}_{3} \]

Key concepts

Reduction Reactions

Reduction reactions involve the addition of hydrogen or the removal of oxygen from a compound. In the context of ethane production, these reactions typically involve reducing agents like hydrogen iodide (HI) and red phosphorus. The reduction of acetic acid \(\mathrm{CH}_{3}\mathrm{COOH}\) using HI and red phosphorus forms ethane \(\mathrm{CH}_{3}\mathrm{CH}_{3}\). However, not all reductions yield ethane. For instance, reducing acetone \(\mathrm{CH}_{3}\mathrm{COCH}_{3}\) with HI and red phosphorus yields propane \(\mathrm{CH}_3 \mathrm{CH}_{2}\mathrm{CH}_3\) instead. It's crucial to identify the starting compound and the specific reagents to predict the products correctly.

Soda Lime Decarboxylation

Soda lime decarboxylation is a chemical reaction where a carboxylic acid salt is heated with soda lime, a mixture of sodium hydroxide (NaOH) and calcium oxide (CaO). This process removes the carboxyl group as carbon dioxide \(\mathrm{CO}_{2}\), resulting in the formation of a hydrocarbon. For example, sodium propionate \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{COONa}\) undergoes soda lime decarboxylation to produce ethane \(\mathrm{CH}_{3}\mathrm{CH}_{3}\). The reaction can be represented as:
\[\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{COONa} + NaOH \xrightarrow{\text{soda lime}} \mathrm{CH}_{3}\mathrm{CH}_{3} + Na_2CO_3\]
This method is commonly used to reduce the carbon chain length of carboxylic acids, producing simpler hydrocarbons.

Hydrogenation Reaction

Hydrogenation reactions involve the addition of hydrogen (H_2) to a molecule, often in the presence of a catalyst. In the case of producing ethane, hydrogenation of ethene \(\mathrm{CH}_{2} = \mathrm{CH}_{2}\) using a Raney nickel catalyst is employed. This process converts the double bond in ethene to a single bond, forming ethane \(\mathrm{CH}_{3}\mathrm{CH}_{3}\). The chemical reaction is:
\[ \mathrm{CH}_{2} = \mathrm{CH}_{2} + H_{2} \xrightarrow{\text{Raney Ni}} \mathrm{CH}_{3}\mathrm{CH}_{3} \]
Hydrogenation is a widely used method in organic chemistry and industrial applications to saturate unsaturated bonds, making the compounds more stable and useful for various purposes.