Identify X in the following reaction. p-Br-C6H4-Cl (1.0 mol) ->[Mg (1.0 mol)][dry ether] Intermediate ->[D2O] X

Correct answer: (C) (C) p-Cl-C6H4-D. The C-Br bond is weaker and more reactive than C-Cl, so with only 1 mol Mg the Grignard forms chemoselectively at the bromine end: p-Br-C6H4-Cl ->[Mg][dry ether] p-ClC6H4-MgBr. The C-Cl bond is untouched. D2O then protonates (deuterates) the C-Mg carbon: p-ClC6H4-MgBr + D2O -> p-Cl-C6H4-D + Mg(OD)Br, giving 4-deuteriochlorobenzene.

NEET 2023 Phase 2 · ChemistryPrevious Year Question

Identify $X$ in the following reaction. $p - Br - C X_{6} H X_{4} - Cl$ (1.0 mol) $Mg (1.0 mol) dry ether$ Intermediate $D X_{2} O$ $X$

Answer: (C) (C) $p - Cl - C X_{6} H X_{4} - D$ . The $C - Br$ bond is weaker and more reactive than $C - Cl$ , so with only 1 mol Mg the Grignard forms chemoselectively at the bromine end: $p - Br - C X_{6} H X_{4} - Cl Mg dry ether p - ClC X_{6} H X_{4} - MgBr$ .

A.(A) $p - D - C X_{6} H X_{4} - D$
B.(B) $p - D - C X_{6} H X_{4} - Br$
C.(C) $p - Cl - C X_{6} H X_{4} - D$ ✓
D.(D) $p - DO - C X_{6} H X_{4} - OD$

Correct Answer

Solution & Explanation

The $C - Br$ bond is weaker and more reactive than $C - Cl$ , so with only 1 mol Mg the Grignard forms chemoselectively at the bromine end: $p - Br - C X_{6} H X_{4} - Cl Mg dry ether p - ClC X_{6} H X_{4} - MgBr$ . The $C - Cl$ bond is untouched. $D X_{2} O$ then protonates (deuterates) the $C - Mg$ carbon: $p - ClC X_{6} H X_{4} - MgBr + D X_{2} O p - Cl - C X_{6} H X_{4} - D + Mg (OD) Br$ , giving 4-deuteriochlorobenzene.

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Identify X in the following reaction. p-Br-CX6​HX4​-Cl (1.0 mol) Mg (1.0 mol)dry ether​ Intermediate DX2​O​ X

Solution & Explanation

More NEET PYQ solutions

Identify $X$ in the following reaction. $p - Br - C X_{6} H X_{4} - Cl$ (1.0 mol) $Mg (1.0 mol) dry ether$ Intermediate $D X_{2} O$ $X$