Question

In the following reaction sequence, \(Q\), \(R\), \(S\) and \(T\) are the major products. The correct statement(s) about \(Q\), \(R\), \(S\) and \(T\) is(are)

• A. \(S\) on warming with ammoniacal \(\mathrm{AgNO_3}\) results in the formation of silver mirror.
• B. \(Q\) on treatment with \(\mathrm{Cl_2}\)(excess)/UV gives gammexane.
• C. \(T\) is a heterocyclic compound.
• D. \(R\) on acid catalyzed intramolecular cyclization followed by treatment with \(\mathrm{Zn-Hg/HCl}\) gives \(9,10\)-dihydroanthracene.

Hint:

Important named reactions used here:
• Kolbe electrolysis: \[ 2RCOO^- \rightarrow R-R + 2CO_2 \]
• Rosenmund reduction: \[ \mathrm{RCOCl \xrightarrow{H_2/Pd-BaSO_4} RCHO} \]
• Wolff–Kishner reduction: \[ \mathrm{C=O \rightarrow CH_2} \]
• Tollens' test: aldehydes give silver mirror. Always identify products step-by-step in multistage organic reaction sequences.

Answer 1

The Correct Option is A

Concept: This problem involves a multistep organic reaction sequence containing:
• Kolbe's electrolysis
• Friedel–Crafts acylation
• Rosenmund reduction
• Wolff–Kishner reduction
• Intramolecular cyclization reactions The strategy is:
• Identify product \(Q\)
• Use \(Q\) to determine product \(R\)
• Follow the subsequent reductions to identify \(S\) and \(T\)
• Finally evaluate each statement carefully

Step 1: Formation of product \(Q\) by Kolbe's electrolysis. The starting compound is sodium propionate: \[ \mathrm{CH_3CH_2COONa} \] In Kolbe electrolysis, carboxylate ions undergo anodic decarboxylation: \[ \mathrm{RCOO^- \rightarrow R^\bullet + CO_2 + e^-} \] For sodium propionate: \[ \mathrm{CH_3CH_2COO^- \rightarrow CH_3CH_2^\bullet + CO_2} \] Two ethyl radicals combine: \[ \mathrm{CH_3CH_2^\bullet + CH_3CH_2^\bullet \rightarrow CH_3CH_2CH_2CH_3} \] Thus product \(Q\) is: \[ Q = n\text{-butane} \]

Step 2: Checking statement (B). Statement (B) says: \[ Q \text{ on treatment with } \mathrm{Cl_2}(\text{excess})/\mathrm{UV} \text{ gives gammexane} \] But: \[ Q = n\text{-butane} \] Gammexane (BHC or benzene hexachloride) is produced from benzene: \[ \mathrm{C_6H_6 + 3Cl_2 \xrightarrow{UV} C_6H_6Cl_6} \] Since butane cannot produce gammexane, statement (B) is incorrect.

Step 3: Formation of product \(R\). Now \(Q\) reacts with phthalic anhydride in presence of anhydrous \(\mathrm{AlCl_3}\). This is a Friedel–Crafts acylation reaction. The alkylbenzene derivative formed undergoes acylation with the phthalic anhydride moiety to produce an aromatic keto-acid derivative \(R\). This product contains both:
• aromatic ring
• ketonic functionality suitable for further transformations

Step 4: Formation of product \(S\). The reagents used are: \[ 1.\ \mathrm{PCl_5} \qquad 2.\ \mathrm{H_2/Pd-BaSO_4} \] This sequence corresponds to:
• conversion of carboxylic acid into acid chloride
• Rosenmund reduction of acid chloride into aldehyde Thus product \(S\) contains an aldehyde group.

Step 5: Checking statement (A). Statement (A) says: \[ S \text{ on warming with ammoniacal } \mathrm{AgNO_3} \text{ gives silver mirror} \] Ammoniacal silver nitrate is Tollens' reagent. Tollens' reagent gives positive test with aldehydes. Since \(S\) contains an aldehyde group after Rosenmund reduction, it will indeed produce silver mirror. Therefore statement (A) is correct.

Step 6: Formation of product \(T\). Now \(S\) is treated with: \[ \mathrm{NH_2NH_2,\ heat} \] This is Wolff–Kishner reduction. The carbonyl group gets reduced to methylene group. During the process, cyclization leads to formation of a ring-containing product. The final structure contains a heterocyclic ring system. Thus: \[ T \text{ is heterocyclic} \] Therefore statement (C) is correct.

Step 7: Checking statement (D). Statement (D) states: \[ R \text{ on acid catalyzed intramolecular cyclization followed by } \mathrm{Zn-Hg/HCl} \] gives: \[ 9,10\text{-dihydroanthracene} \] The structure of \(R\) indeed possesses the appropriate arrangement for intramolecular Friedel–Crafts cyclization. Cyclization produces anthracene-type fused ring ketone intermediate. Subsequent Clemmensen reduction: \[ \mathrm{Zn-Hg/HCl} \] reduces the carbonyl group to methylene group yielding: \[ 9,10\text{-dihydroanthracene} \] Hence statement (D) is correct.

Step 8: Final conclusion. Correct statements are: \[ (A),\ (C)\ \text{and}\ (D) \] Final Answer: \[ \boxed{(A),\ (C)\ \text{and}\ (D)} \]