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

Step 1: Read the clues backward.
$S$ is a hydrocarbon made from $Q$ by soda lime. Soda lime removes one carbon as carbonate, so $Q$ must be a carboxylic acid and $S$ has one carbon less.

Step 2: Use the Grignard clue.
$P$ plus ethyl magnesium bromide, then acid, gives an acid $Q$. A Grignard reagent adds to carbon dioxide and after hydrolysis gives a carboxylic acid with one extra carbon. So $P$ is $CO_2$.

Step 3: Build Q.
Ethyl group ($C_2H_5$) joining $CO_2$ gives $C_2H_5COOH$, that is propanoic acid. So $Q = C_2H_5COOH$.

Step 4: Apply the HVZ reaction.
$Q$ with $Cl_2$ and red phosphorus undergoes alpha halogenation (Hell-Volhard-Zelinsky), putting $Cl$ on the carbon next to $COOH$. This gives $R = CH_3CHClCOOH$.

Step 5: Apply decarboxylation.
$Q$ with soda lime loses $CO_2$, leaving the hydrocarbon with one carbon less than the acid. $C_2H_5COOH$ has three carbons, so $S$ has two carbons, giving ethane, $C_2H_6$.

Step 6: Collect the answer.
So $P, Q, R, S$ are $CO_2$, $C_2H_5COOH$, $CH_3CHClCOOH$, $C_2H_6$.
\[ \boxed{CO_2,\ C_2H_5COOH,\ CH_3CHClCOOH,\ C_2H_6} \]