Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is conver

Question

Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to a gaseous hydrocarbon containing less than four carbon atoms. The formula of (A) is:

Accepted Answer

1. **Analyze Reaction 1 (Substitution):** The hydrocarbon (A) reacts with bromine via a **substitution** reaction. This indicates that (A) is a saturated hydrocarbon (Alkane), as unsaturated hydrocarbons (Alkenes/Alkynes) typically undergo addition reactions with halogens [NCERT 11th, Sec 9.2.3]. This eliminates Options A ($CH \equiv CH$) and B ($CH_2=CH_2$).
2. **Analyze Reaction 2 (Wurtz Reaction):** The alkyl bromide formed reacts with sodium (Wurtz reaction) to double its carbon chain.
   $$ 2R-Br + 2Na \xrightarrow{	ext{dry ether}} R-R + 2NaBr $$
3. **Analyze Product Constraints:** The final product is a **gaseous hydrocarbon** with **less than four** carbon atoms.
   - **Case 1: If (A) is Methane ($CH_4$):**
     - Bromination: $CH_4 + Br_2 \xrightarrow{h
u} CH_3Br$ (Methyl bromide).
     - Wurtz Reaction: $2CH_3Br + 2Na ightarrow CH_3-CH_3$ (Ethane).
     - Ethane ($C_2H_6$) is a gas and has 2 carbon atoms (which is less than 4). **This fits the criteria.**
   - **Case 2: If (A) is Ethane ($CH_3-CH_3$):**
     - Bromination: $C_2H_6 + Br_2 ightarrow C_2H_5Br$ (Ethyl bromide).
     - Wurtz Reaction: $2C_2H_5Br + 2Na ightarrow C_4H_{10}$ (Butane).
     - Butane ($C_4H_{10}$) has 4 carbon atoms. The condition requires *less than* four. **This does not fit.**
4. **Conclusion:** The starting hydrocarbon (A) must be Methane ($CH_4$).

Step-by-Step Solution

Practice All CHEMISTRY Questions