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Step-by-Step Solution
Step 1: Understanding the Reaction with Bromine
Hydrocarbons that undergo substitution reactions with bromine (Br2) in the presence of suitable conditions (like UV light) are generally alkanes. Thus, for the given hydrocarbon (A), the reaction leads to the formation of an alkyl bromide.
Step 2: Applying the Wurtz Reaction
The Wurtz reaction involves the treatment of an alkyl halide with sodium metal in dry ether to form a new carbon–carbon bond. The general form of the Wurtz reaction is:
$ 2 \, R-X + 2 \, Na \longrightarrow R-R + 2 \, NaX $
So, if (A) first gives an alkyl bromide (R–Br), then upon Wurtz reaction, we obtain an alkane with fewer than four carbon atoms in total (suggesting it is ethane or propane).
Step 3: Deductions Leading to Methane (CH4)
1. If (A) were an alkene (CH2=CH2) or alkyne (CH≡CH), they would generally react with bromine via addition rather than substitution.
2. If (A) were ethane (CH3-CH3), the Wurtz reaction product would have more carbons than two (specifically butane).
3. The only choice that satisfies bromination by substitution (forming CH3Br) and then yields a product containing fewer than four carbons after Wurtz reaction (namely ethane, which has two carbons) is methane (CH4).
Therefore, (A) must be CH4.
