Question

The empirical formula of the compound 'D' formed in the given reaction sequence is
C₂H₄ \(\xrightarrow{Br_2|CCl_4}\) A \(\xrightarrow{(i) alc. KOH (ii) NaNH_2}\) B \(\xrightarrow{cyclic\;polymerization}\) C \(\xrightarrow{Cl_2(excess), dry\;AlCl_3, dark, cold}\) D

• A. CHCl
• B. CCl
• C. CH₂Cl
• D. CHCl₂

Hint:

Remember these key transformations: 1. Alkene + Halogen \(\rightarrow\) Vicinal Dihalide. 2. Vicinal Dihalide + alc. KOH/NaNH₂ \(\rightarrow\) Alkyne. 3. Alkyne (Ethyne) + Red hot Fe tube \(\rightarrow\) Benzene. 4. Benzene + Excess Halogen/Catalyst \(\rightarrow\) Hexa-substituted Benzene.

Answer 1

The Correct Option is B

Step 1: Reactant to A: \( \text{C}_2\text{H}_4 \) (Ethene) + \( \text{Br}_2/\text{CCl}_4 \) \( \to \) \( \text{Br-CH}_2-\text{CH}_2-\text{Br} \) (1,2-Dibromoethane). A = 1,2-Dibromoethane.
Step 2: A to B: 1,2-Dibromoethane + (i) alc. KOH + (ii) \( \text{NaNH}_2 \) (Strong base, dehydrohalogenation twice). \( \to \) Acetylene (\( \text{HC}\equiv\text{CH} \)). B = Ethyne (\( \text{C}_2\text{H}_2 \)).
Step 3: B to C: Ethyne \( \xrightarrow{\text{cyclic polymerization}} \) Benzene (\( \text{C}_6\text{H}_6 \)). C = Benzene.
Step 4: C to D: Benzene + \( \text{Cl}_2 \) (excess) + dry \( \text{AlCl}_3 \) (dark, cold). This implies electrophilic substitution (Chlorination) on the ring. Since \( \text{Cl}_2 \) is in excess and \( \text{AlCl}_3 \) is present, all hydrogen atoms on the benzene ring are replaced by chlorine atoms. Reaction: \( \text{C}_6\text{H}_6 + 6\text{Cl}_2 \xrightarrow{\text{AlCl}_3} \text{C}_6\text{Cl}_6 + 6\text{HCl} \). Product D is Hexachlorobenzene (\( \text{C}_6\text{Cl}_6 \)). Note: If it were sunlight/UV, it would be addition (BHC - C6H6Cl6). With Lewis acid, it is substitution.
Step 5: Empirical Formula of D: Molecular Formula of D = \( \text{C}_6\text{Cl}_6 \). Ratio of C:Cl = 6:6 = 1:1. Empirical Formula = \( \text{CCl} \).