The Question is from neet-chemistry hydrocarbons , and it was created by the author Mr. S M HUSSAIN

Question

An alkane (A) undergoes a substitution reaction with bromine to yield an alkyl bromide. This alkyl bromide is then subjected to the Wurtz reaction, producing a gaseous hydrocarbon with a molecular mass less than 30. What is the identity of alkane (A)?

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A. Methane (CH₄)
## CONCEPT FRAMEWORK
	**Primary Concept**: Alkane Reactions (Substitution and Wurtz Reaction)
	**Related Topics**: Halogenation of Alkanes, Coupling Reactions, Molecular Mass Calculation
	**Prerequisites**: Understanding of alkane reactivity, Wurtz reaction mechanism, and basic stoichiometry.
## SOLUTION PATHWAY
**Given Information**:
	- Alkane (A) reacts with Bromine (Br₂) in a substitution reaction.
	- The resulting alkyl bromide undergoes a Wurtz reaction.
	- The final product is a gaseous hydrocarbon with a molecular mass less than 30.
**Method & Steps**:
	1. **Analyze the Wurtz Reaction**: The Wurtz reaction involves the coupling of two alkyl halides in the presence of sodium metal to form a higher alkane.  The general reaction is: 2R-X + 2Na → R-R + 2NaX
		Reasoning: This implies the final alkane will have double the number of carbon atoms of the initial alkyl halide (excluding Methane which forms Ethane, still gaseous).
	2. **Consider the Molecular Mass Constraint**: The product alkane has a molecular mass less than 30.
		Working: Ethane (C₂H₆) has a molecular mass of 30.  Propane (C₃H₈) is 44, Butane (C₄H₁₀) is 58 and so on.
		Key Point:  This significantly limits the possibilities for the product and, consequently, the reactant alkyl bromide.
	3. **Determine Alkane (A)**: Since the product must be a gas with a molecular mass less than 30 and Ethane is 30, it follows that the final product must be Ethane. If the product of the Wurtz reaction is ethane, then the alkyl halide must be methyl bromide (CH₃Br). Methyl bromide is formed by the substitution reaction of methane (CH₄) with bromine.
		Result: Alkane (A) is Methane (CH₄).
		Verification: CH₄ + Br₂ → CH₃Br + HBr;  2CH₃Br + 2Na → C₂H₆ + 2NaBr. Ethane has a molar mass of 30.
## OPTION ANALYSIS
[A] ✓: Methane (CH₄) is correct. As explained above, methane reacts with bromine to form methyl bromide, which then undergoes the Wurtz reaction to yield ethane, a gas with a molecular mass less than 30 (it is 30, but all other options produce gases with higher molecular masses). 
[B] :x: Ethane (C₂H₆): If ethane were alkane (A), the Wurtz reaction would produce butane (C₄H₁₀), which has a molecular mass of 58, exceeding the limit of 30.
[C] :x: Propane (C₃H₈):  If propane were alkane (A), the Wurtz product would be hexane (C₆H₁₄), with a molecular mass much greater than 30.
[D] :x: Butane (C₄H₁₀): If butane were alkane (A), the product would be octane (C₈H₁₈), again with a molecular mass far exceeding 30.
## LEARNING AIDS
	**Common Mistakes**:
		1. Ignoring the molecular mass constraint:  Students might overlook the importance of the molecular mass limit and incorrectly choose a larger alkane.
			How to avoid: Pay close attention to all constraints provided in the problem.
		2. Misinterpreting the Wurtz Reaction: Not realizing that the product alkane has double the number of carbons (except in the case of Methane) of the alkyl halide can lead to incorrect choices.
			How to avoid: Review the mechanism and general equation of the Wurtz reaction.
	**Quick Tips**:
		1. Work backward from the product: Knowing the product has a mass less than 30 allows you to quickly narrow down the possibilities for the alkyl halide and then the starting alkane.
		2. Remember the Wurtz Reaction pattern:  2R-X → R-R. This helps in quickly determining the relationship between the alkyl halide and the final product.

Mr. S M HUSSAIN · Answer

The correct Option is "Methane (CH₄)"

Mr. S M HUSSAIN · Answer

Ethane (C₂H₆)

Mr. S M HUSSAIN · Answer

Propane (C₃H₈)

Mr. S M HUSSAIN · Answer

Butane (C₄H₁₀)

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