In the following reaction sequence, Q, R, S and T are the major products. [figure] The correct statement(s) about Q, R, S and T is(are)
- S on warming with ammoniacal AgNO3 results in the formation of silver mirror.
- Q on treatment with Cl2(excess)/UV gives gammaxane.
- T is a heterocyclic compound.
- R on acid catalyzed intramolecular cyclization followed by treatment with Zn-Hg/HCl gives 9,10-dihydroxyanthracene.
Answer
(1) S on warming with ammoniacal AgNO3 results in the formation of silver mirror., (2) Q on treatment with Cl2(excess)/UV gives gammaxane., (3) T is a heterocyclic compound.
Solution
Kolbe electrolysis: two propyl radicals (from 2 RCOO⁻, –2CO₂) couple: [figure] electrolysis [figure]
Aromatisation (cyclisation + dehydrogenation): [figure] [ 10-20 atm]V2O5, 773 K [figure] = Q
Friedel–Crafts acylation (anhydride opens, one C=O acylates benzene): [figure] + [figure] anhyd. AlCl3 [figure] = R
–COOH → –COCl, then Rosenmund reduction –COCl → –CHO: [figure] PCl5 [figure] H2-Pd/BaSO4 [figure] = S
Both C=O groups (1,4-related) condense with the two –NH₂ of hydrazine: [figure] [-2H2O]NH2NH2, heat [figure] = T (1-phenylphthalazine)
(A) S has a –CHO group, so it gives a silver mirror with Tollens' reagent. Correct.
(B) C6H6 + 3Cl2 UV C6H6Cl6 (BHC, gammaxane). Correct.
(C) T has an N–N containing ring, so it is heterocyclic. Correct.
(D) [figure] [-H2O] H+ [figure] Zn-Hg/HCl [figure]
Clemmensen reduction removes the C=O groups (gives 9,10-dihydroanthracene), not 9,10-dihydroxyanthracene. Wrong.
Answer: (A), (B), (C)