Understanding the Reaction Sequence

The given problem involves a sequence of organic reactions starting from acetic acid. We need to follow the reactions step-by-step to determine the final product 'C' and then identify a homologue of 'C' from the given options.

The reaction sequence is:

\({\rm{C}}{{\rm{H}}_3}{\rm{COOH}}\mathop \to \limits^{LiAI{H_4}} {\rm{A}}\mathop \to \limits^{PC{I_5}} {\rm{B}}\mathop \to \limits^{alc.KOH} {\rm{C}}\)

Step 1: Reduction of Acetic Acid

The first step involves the reaction of acetic acid \({\rm{C}}{{\rm{H}}_3}{\rm{COOH}}\) with Lithium Aluminum Hydride (\({\rm{LiAI{H_4}}}\)). \({\rm{LiAI{H_4}}}\) is a powerful reducing agent that reduces carboxylic acids to primary alcohols.

The reaction is:

\({\rm{C}}{{\rm{H}}_3}{\rm{COOH}} \xrightarrow{LiAI{H_4}} {\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{OH}}\)

So, product 'A' is ethanol \({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{OH}}\).

Step 2: Reaction of Alcohol with \({\rm{PC{l_5}}}\)

The second step is the reaction of ethanol 'A' with Phosphorus Pentachloride (\({\rm{PC{l_5}}}\)). \({\rm{PC{l_5}}}\) is a chlorinating agent that replaces the hydroxyl (-OH) group in alcohols with a chlorine (-Cl) atom, forming an alkyl halide.

The reaction is:

\({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{OH}} + {\rm{PC{l_5}}} \to {\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{Cl}} + {\rm{POC{l_3}}} + {\rm{HCl}}\)

So, product 'B' is chloroethane \({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{Cl}}\).

Step 3: Dehydrohalogenation with Alcoholic \({\rm{KOH}}\)

The third step is the reaction of chloroethane 'B' with alcoholic Potassium Hydroxide (\({\rm{alc. KOH}}\)). Alcoholic \({\rm{KOH}}\) is used for dehydrohalogenation, which is an elimination reaction where a hydrogen atom from one carbon and a halogen atom from an adjacent carbon are removed, forming an alkene.

The reaction is:

\({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{Cl}} \xrightarrow{alc. KOH}} {\rm{C}}{{\rm{H}}_2}{\rm{=}}{{\rm{CH}}_2} + {\rm{KCl}} + {{\rm{H}}_2}{\rm{O}}\)

So, product 'C' is ethene \({\rm{C}}{{\rm{H}}_2}{\rm{=}}{{\rm{CH}}_2}\), which has the molecular formula \({\rm{C}}{{\rm{H}}_4}\).

Identifying the Homologue of ‘C’ (Ethene)

The final product 'C' is ethene \({\rm{C}}{{\rm{H}}_2}{\rm{=}}{{\rm{CH}}_2}\) with the molecular formula \({\rm{C}}{{\rm{H}}_4}\). The question asks for a homologue of 'C'.

A homologous series is a series of compounds with the same general formula, usually varying by a repeating unit like \({\rm{-C}}{{\rm{H}}_2}{\rm{-}}\). Members of a homologous series are called homologues and have similar chemical properties.

Alkenes form a homologous series with the general formula \({\rm{C}}_n{\rm{H}}_{2n}\). The series starts with:

• Ethene (n=2): \({\rm{C}}_2{\rm{H}}_{2 \times 2} = {\rm{C}}_2{\rm{H}}_4\)
• Propene (n=3): \({\rm{C}}_3{\rm{H}}_{2 \times 3} = {\rm{C}}_3{\rm{H}}_6\)
• Butene (n=4): \({\rm{C}}_4{\rm{H}}_{2 \times 4} = {\rm{C}}_4{\rm{H}}_8\)
• and so on...

Homologues of ethene (\({\rm{C}}_2{\rm{H}}_4\)) would typically be other alkenes like propene (\({\rm{C}}_3{\rm{H}}_6\)), butene (\({\rm{C}}_4{\rm{H}}_8\)), etc., which differ by a \({\rm{-C}}{{\rm{H}}_2}{\rm{-}}\) group.

Analyzing the Options

Let's examine the given options:

• Option 1: \({\rm{C}}_3{\rm{H}}_6\) - This is propene. It fits the general formula \({\rm{C}}_n{\rm{H}}_{2n}\) for n=3. It is a homologue of ethene, differing by a \({\rm{-C}}{{\rm{H}}_2}{\rm{-}}\) group.
• Option 2: \({\rm{C}}_2{\rm{H}}_4{\rm{CHO}}\) - This formula corresponds to \({\rm{C}}_3{\rm{H}}_5{\rm{O}}\) (if it means \({\rm{C_2H_4}} + \text{CHO}\)) or it might represent propanal \({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{CHO}}\) which is \({\rm{C}}_3{\rm{H}}_6{\rm{O}}\). Neither interpretation represents an alkene or a homologue of ethene.
• Option 3: \({\rm{C}}_2{\rm{H}}_4\) - This is ethene itself. It is the compound 'C' that was formed in the reaction.
• Option 4: \({\rm{C}}_2{\rm{H}}_2\) - This is ethyne (acetylene). It fits the general formula \({\rm{C}}_n{\rm{H}}_{2n-2}\) for alkynes. It belongs to a different homologous series than alkenes.

Based on the reaction sequence, 'C' is \({\rm{C}}_2{\rm{H}}_4\). Among the provided options, option 3 is \({\rm{C}}_2{\rm{H}}_4\).

The final product 'C' is ethene (\({\rm{C}}_2{\rm{H}}_4\)). The question asks for a homologue of 'C'. Option 3 is \({\rm{C}}_2{\rm{H}}_4\), which is 'C' itself.

Conclusion

The product 'C' obtained from the reaction sequence is ethene (\({\rm{C}}_2{\rm{H}}_4}\)). Among the given options, option 3 matches the formula of ethene, \({\rm{C}}_2{\rm{H}}_4}\).

Revision Table: Key Reaction Types

Additional Information: Homologous Series

A homologous series is a series of organic compounds that have the same functional group and similar chemical properties. Each successive member in the series differs by a constant group, usually a \({\rm{-C}}{{\rm{H}}_2}{\rm{-}}\) unit.

Examples of homologous series include:

• Alkanes: \({\rm{CH}}_4\), \({\rm{C}}_2{\rm{H}}_6\), \({\rm{C}}_3{\rm{H}}_8\), ... (General formula: \({\rm{C}}_n{\rm{H}}_{2n+2}\))
• Alkenes: \({\rm{C}}_2{\rm{H}}_4\), \({\rm{C}}_3{\rm{H}}_6\), \({\rm{C}}_4{\rm{H}}_8\), ... (General formula: \({\rm{C}}_n{\rm{H}}_{2n}\))
• Alkynes: \({\rm{C}}_2{\rm{H}}_2\), \({\rm{C}}_3{\rm{H}}_4\), \({\rm{C}}_4{\rm{H}}_6\), ... (General formula: \({\rm{C}}_n{\rm{H}}_{2n-2}\))
• Alcohols: \({\rm{CH}}_3{\rm{OH}}\), \({\rm{C}}_2{\rm{H}}_5{\rm{OH}}\), \({\rm{C}}_3{\rm{H}}_7{\rm{OH}}\), ... (General formula: \({\rm{C}}_n{\rm{H}}_{2n+1}{\rm{OH}}\))

In this question, 'C' is ethene (\({\rm{C}}_2{\rm{H}}_4}\)), which is the first member of the alkene homologous series. Propene (\({\rm{C}}_3{\rm{H}}_6}\)) would be the next homologue in this series.