Question

A compound \(P\) on treatment with \(C_2H_5MgBr\) in presence of ether followed by acidic hydrolysis produces a compound \(Q\). \(Q\) on treatment with \(Cl_2/red\;P\) in mild condition produces a compound \(R\). Compound \(Q\) on heating with soda lime produces a hydrocarbon \(S\). Compounds \(P\), \(Q\), \(R\) and \(S\) are respectively

• A. \(HCHO,\; C_2H_5COOH,\; CH_3CHClCOOH,\; CH_4\)
• B. \(HCHO,\; CH_3COOH,\; CH_3COCl,\; C_2H_6\)
• C. \(CO_2,\; C_2H_5COOH,\; CH_3CHClCOOH,\; C_2H_6\)
• D. \(CO_2,\; CH_3COOH,\; CH_3COCl,\; CH_4\)

Hint:

Important reactions: \[ RMgX + CO_2 \xrightarrow{H^+} RCOOH \] \[ RCH_2COOH \xrightarrow{Cl_2/red\,P} RCHClCOOH \] \[ RCOOH \xrightarrow{\text{soda lime}} RH \] Remember: Grignard reagent + \(CO_2\) increases the carbon chain by one carbon atom.

Answer 1

The Correct Option is C

Concept: Grignard reagents react with carbon dioxide to form carboxylic acids after acidic hydrolysis. \[\begin{aligned} RMgX + CO_2 \rightarrow RCOOMgX \xrightarrow{H^+} RCOOH \end{aligned}\] Carboxylic acids undergo Hell-Volhard-Zelinsky (HVZ) reaction with \(Cl_2/red\;P\) to form \(\alpha\)-halo acids. Soda-lime decarboxylation gives an alkane having one carbon atom less than the parent acid.

Step 1: Identify compound \(P\). Ethyl magnesium bromide reacts with carbon dioxide as \[\begin{aligned} C_2H_5MgBr + CO_2 \rightarrow C_2H_5COOMgBr \end{aligned}\] On acidic hydrolysis, \[\begin{aligned} C_2H_5COOMgBr \xrightarrow{H^+} C_2H_5COOH \end{aligned}\] Thus, \[\begin{aligned} P=CO_2,\qquad Q=C_2H_5COOH \end{aligned}\]

Step 2: Identify compound \(R\). Propanoic acid undergoes HVZ reaction. \[\begin{aligned} CH_3CH_2COOH \xrightarrow{Cl_2/red\,P} CH_3CHClCOOH \end{aligned}\] Therefore, \[\begin{aligned} R=CH_3CHClCOOH \end{aligned}\]

Step 3: Identify compound \(S\). Soda-lime decarboxylation: \[\begin{aligned} CH_3CH_2COOH \xrightarrow{\text{soda lime}} C_2H_6 \end{aligned}\] Thus, \[\begin{aligned} S=C_2H_6 \end{aligned}\]

Step 4: Summarize the compounds. \[ \begin{aligned} P &:\quad CO_2 \\ Q &:\quad C_2H_5COOH \\ R &:\quad CH_3CHClCOOH \\ S &:\quad C_2H_6 \end{aligned} \] \[\begin{aligned} \boxed{ CO_2,\; C_2H_5COOH,\; CH_3CHClCOOH,\; C_2H_6 } \end{aligned}\] Hence, option \(\mathbf{(C)}\) is correct.