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NEET CHEMISTRYMedium

The compound A on treatment with Na gives B, and with PCl₅ gives C. B and C react together to give diethyl ether. A, B and C are, respectively:

A

C2H5OH,C2H6,C2H5ClC_2H_5OH, C_2H_6, C_2H_5Cl

B

C2H5OH,C2H5Cl,C2H5ONaC_2H_5OH, C_2H_5Cl, C_2H_5ONa

C

C2H5Cl,C2H6,C2H5OHC_2H_5Cl, C_2H_6, C_2H_5OH

D

C2H5OH,C2H5ONa,C2H5ClC_2H_5OH, C_2H_5ONa, C_2H_5Cl

Step-by-Step Solution

  1. Identify A: Since the final product is diethyl ether (C2H5−O−C2H5C_2H_5-O-C_2H_5) formed by the reaction of B and C (Williamson synthesis), and A yields B with Na and C with PCl5PCl_5, A must be Ethanol (C2H5OHC_2H_5OH).
  2. Formation of B: Alcohols react with active metals like sodium to form sodium alkoxides. 2C2H5OH(A)+2Na→2C2H5ONa(B)+H22C_2H_5OH (A) + 2Na \rightarrow 2C_2H_5ONa (B) + H_2 So, B is Sodium Ethoxide.
  3. Formation of C: Alcohols react with phosphorus pentachloride (PCl5PCl_5) to form alkyl chlorides. C2H5OH(A)+PCl5→C2H5Cl(C)+POCl3+HClC_2H_5OH (A) + PCl_5 \rightarrow C_2H_5Cl (C) + POCl_3 + HCl So, C is Ethyl Chloride.
  4. Formation of Ether: B and C react via nucleophilic substitution (Williamson Ether Synthesis). C2H5ONa(B)+C2H5Cl(C)→C2H5−O−C2H5+NaClC_2H_5ONa (B) + C_2H_5Cl (C) \rightarrow C_2H_5-O-C_2H_5 + NaCl
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