Question

Observed the following reaction sequence and identify the correct options?

• A. F reacts with $KMnO_4$ / $H^+$ gives phthalic acid which upon heating gives phthalic anhydride.
• B. E upon monochlorination ($Cl_2$ / hv) gives total four products.
• C. F upon complete catalytic hydrogenation gives saturated compound which show geometrical isomerism.
• D. Compound (A) gives positive test with $NaHCO_3$.

Answer 1

Answer: (A), (B), (C), (D)

A, B, C, D

Answer 2

1. Benzene + Succinic anhydride $\xrightarrow{AlCl_3}$ (A): Friedel-Crafts acylation yields 4-oxo-4-phenylbutanoic acid (A).
2. $A$ $\xrightarrow{Zn-Hg/HCl}$ (B): Clemmensen reduction converts the ketone to a methylene group, yielding 4-phenylbutanoic acid (B).
3. $B$ $\xrightarrow{SOCl_2}$ (C): Thionyl chloride converts the carboxylic acid to an acyl chloride, yielding 4-phenylbutanoyl chloride (C).
4. $C$ $\xrightarrow{AlCl_3}$ (D): Intramolecular Friedel-Crafts acylation forms a 6-membered ring, yielding 1-tetralone (D).
5. $D$ $\xrightarrow{Zn-Hg/HCl}$ (E): Clemmensen reduction of the ketone yields tetralin (E).
6. $E$ $\xrightarrow{Pd-C/\Delta}$ (F): Catalytic dehydrogenation of tetralin yields naphthalene (F).

Evaluating options: (A) Naphthalene (F) oxidation yields phthalic acid, which dehydrates to phthalic anhydride upon heating. Correct. (B) Tetralin (E) has 4 types of non-equivalent hydrogens (2 on the aromatic ring, 2 on the saturated ring - benzylic and non-benzylic), leading to 4 monochlorination products. Correct. (C) Complete hydrogenation of naphthalene (F) gives decalin, which exists as cis and trans isomers (geometrical isomerism). Correct. (D) 4-oxo-4-phenylbutanoic acid (A) has a carboxylic acid group, which reacts with $NaHCO_3$ to give a positive test. Correct.