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: Understanding the Concept:
This is a multi-step organic synthesis problem. We need to identify the product at each stage of the reaction sequence to determine the final product 'D' and then find its empirical formula.
Step 2: Detailed Reaction Sequence:
- Step 1: Ethene (C₂H₄) to A
\[ \text{CH}_2=\text{CH}_2 + \text{Br}_2 \xrightarrow{\text{CCl}_4} \text{Br}-\text{CH}_2-\text{CH}_2-\text{Br} \] Ethene undergoes electrophilic addition with bromine in an inert solvent (CCl₄) to form 1,2-dibromoethane. So, A is 1,2-dibromoethane.
- Step 2: A to B
\[ \text{Br}-\text{CH}_2-\text{CH}_2-\text{Br} \xrightarrow{\text{(i) alc. KOH}} \text{CH}_2=\text{CH}-\text{Br} \xrightarrow{\text{(ii) NaNH}_2} \text{HC}\equiv\text{CH} \] This is a double dehydrohalogenation. Alcoholic KOH (a moderate base) removes one HBr to form vinyl bromide. Sodamide (NaNH₂, a very strong base) is required to remove the second HBr from the less reactive vinyl halide to form ethyne (acetylene). So, B is ethyne.
- Step 3: B to C
\[ 3 \text{ HC}\equiv\text{CH} \xrightarrow{\text{Red hot Fe tube}} \text{C}_6\text{H}_6 \] Ethyne undergoes cyclic polymerization (trimerization) when passed through a red-hot iron tube at high temperature to form benzene. So, C is benzene.
- Step 4: C to D
\[ \text{C}_6\text{H}_6 + 6 \text{Cl}_2 (\text{excess}) \xrightarrow{\text{dry AlCl}_3, \text{dark, cold}} \text{C}_6\text{Cl}_6 + 6 \text{HCl} \] Benzene undergoes electrophilic aromatic substitution (chlorination) in the presence of a Lewis acid catalyst (AlCl₃). Since excess chlorine is used, all six hydrogen atoms on the benzene ring are substituted by chlorine atoms to form hexachlorobenzene. So, D is hexachlorobenzene.
Step 3: Finding the Empirical Formula of D:
The molecular formula of the final product D (hexachlorobenzene) is C₆Cl₆.
The empirical formula is the simplest whole-number ratio of atoms in the compound. To find it, we divide the subscripts by their greatest common divisor, which is 6.
\[ \text{C}_{6/6}\text{Cl}_{6/6} \rightarrow \text{C}_1\text{Cl}_1 \rightarrow \text{CCl} \] Step 4: Final Answer:
The empirical formula of compound D is CCl. Therefore, option (B) is correct.