Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

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:

$CH \equiv CH$

$CH_2 = CH_2$

$CH_3 - CH_3$

$CH_4$

Step-by-Step Solution

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 ( $CH_2=CH_2$ ).
Analyze Reaction 2 (Wurtz Reaction): The alkyl bromide formed reacts with sodium (Wurtz reaction) to double its carbon chain. $2R-Br + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaBr$
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: $CH_4 + Br_2 \xrightarrow{h\nu} CH_3Br$ (Methyl bromide).
Wurtz Reaction: $2CH_3Br + 2Na \rightarrow CH_3-CH_3$ (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: $C_2H_6 + Br_2 \rightarrow C_2H_5Br$ (Ethyl bromide).
Wurtz Reaction: $2C_2H_5Br + 2Na \rightarrow C_4H_{10}$ (Butane).
Butane ( $C_4H_{10}$ ) has 4 carbon atoms. The condition requires less than four. This does not fit.

Conclusion: The starting hydrocarbon (A) must be Methane ( $CH_4$ ).

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