Identify the final product in the following sequence: C6H5N2^+ Cl^- ->[CuBr/HBr] (i) ->[Mg/dry ether] (ii) ->[H2O] Product

Correct answer: (B) (B) Benzene (C6H6). Sandmeyer-type treatment converts benzenediazonium chloride to bromobenzene: C6H5N2^+Cl^- ->[CuBr/HBr] C6H5Br. Reaction with Mg in dry ether gives the Grignard reagent C6H5MgBr, which on contact with water (a source of H) is protonated: C6H5MgBr + H2O -> C6H6 + Mg(OH)Br. The final product is benzene.

NEET 2023 Phase 1 · ChemistryPrevious Year Question

Identify the final product in the following sequence: $C X_{6} H X_{5} N X_{2} X^{+} Cl X^{-} CuBr / HBr (i) Mg / dry ether (ii) H X_{2} O Product$

Answer: (B) (B) Benzene $(C X_{6} H X_{6})$ . Sandmeyer-type treatment converts benzenediazonium chloride to bromobenzene: $C X_{6} H X_{5} N X_{2} X^{+} Cl X^{-} CuBr / HBr C X_{6} H X_{5} Br$ .

A.(A) Phenol $(C X_{6} H X_{5} OH)$
B.(B) Benzene $(C X_{6} H X_{6})$ ✓
C.(C) Phenylmagnesium bromide $(C X_{6} H X_{5} MgBr)$
D.(D) p-Bromophenol

Correct Answer

(B) (B) Benzene $(C X_{6} H X_{6})$

Solution & Explanation

Sandmeyer-type treatment converts benzenediazonium chloride to bromobenzene: $C X_{6} H X_{5} N X_{2} X^{+} Cl X^{-} CuBr / HBr C X_{6} H X_{5} Br$ . Reaction with $Mg$ in dry ether gives the Grignard reagent $C X_{6} H X_{5} MgBr$ , which on contact with water (a source of H) is protonated: $C X_{6} H X_{5} MgBr + H X_{2} O C X_{6} H X_{6} + Mg (OH) Br$ . The final product is benzene.

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Identify the final product in the following sequence: CX6​HX5​NX2​X+ ClX−CuBr/HBr​(i)Mg/dry ether​(ii)HX2​O​Product

Solution & Explanation

More NEET PYQ solutions

Identify the final product in the following sequence: $C X_{6} H X_{5} N X_{2} X^{+} Cl X^{-} CuBr / HBr (i) Mg / dry ether (ii) H X_{2} O Product$