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)
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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.
\(S\) on warming with ammoniacal \(\mathrm{AgNO_3}\) results in the formation of silver mirror.
\(Q\) on treatment with \(\mathrm{Cl_2}\)(excess)/UV gives gammexane.
\(T\) is a heterocyclic compound.
\(R\) on acid catalyzed intramolecular cyclization followed by treatment with \(\mathrm{Zn-Hg/HCl}\) gives \(9,10\)-dihydroanthracene.
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The Correct Option isA
Solution and Explanation
Step 1: Understanding the Concept:
Identify the intermediates in the sequence:
1. \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COONa}\) \(\xrightarrow{\text{Kolbe's Electrolysis}}\) \(\text{n-hexane}\) (\(\text{CH}_3(\text{CH}_2)_4\text{CH}_3\)).
Actually, R-COONa gives R-R. Propyl radical gives n-hexane.
2. \(\text{n-Hexane}\) \(\xrightarrow{\text{V}_2\text{O}_5, 500^{\circ}\text{C}, 10-20 \text{ atm}}\) \(\text{Benzene}\) (\(\mathbf{Q}\)).
This is the catalytic aromatization process.
3. \(\text{Benzene (Q) + Phthalic anhydride} \xrightarrow{\text{Anhyd. AlCl}_3}\) \(\text{o-Benzoylbenzoic acid}\) (\(\mathbf{R}\)). (Friedel-Crafts Acylation).
4. \(\mathbf{R} \xrightarrow{\text{PCl}_5}\) Acid Chloride, then \(\xrightarrow{\text{H}_2\text{-Pd/BaSO}_4}\) (Rosenmund Reduction) \(\rightarrow\) \(\mathbf{S}\).
\(\mathbf{S}\) is the corresponding aldehyde: \(\text{o-benzoylbenzaldehyde}\).
5. \(\mathbf{S} \xrightarrow{\text{NH}_2\text{NH}_2, \text{heat}}\) \(\mathbf{T}\).
Reaction of a 1,4-dicarbonyl equivalent (like o-benzoylbenzaldehyde) with hydrazine usually leads to cyclization to form a heterocyclic ring (Phthalazine derivative). Step 3: Detailed Explanation:
(A) S is an aldehyde. Aldehydes react with Tollen's reagent (ammoniacal silver nitrate) to give a silver mirror. Statement (A) is True.
(B) Q is Benzene. Treatment of Benzene with excess Chlorine in the presence of UV light (radical addition) gives Benzene hexachloride (BHC), commonly known as gammaxane or Lindane. Statement (B) is True.
(C) T is formed by the reaction of hydrazine with the dicarbonyl species. This results in the formation of a nitrogen-containing ring (phthalazine), which is heterocyclic. Statement (C) is True.
(D) R is o-benzoylbenzoic acid. Acid catalyzed cyclization (with H\(_2\)SO\(_4\)) gives Anthraquinone. Treatment of anthraquinone with Zn-Hg/HCl (Clemmensen Reduction) would ideally reduce the carbonyls to \(-CH_2-\) groups (Anthracene), not dihydroxyanthracene. Statement (D) is False. Step 4: Final Answer:
The sequence goes from an aliphatic salt to benzene, then to a keto-acid, aldehyde, and finally a heterocyclic compound.
