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Question

Which among the following molecular formulae of organic compounds does NOT represent a ring structure?

This question was previously asked in
RRB Group D 2025 Question Paper (18-Aug-2026) (Shift 1)
The correct answer is

C6H14

A quick way to tell whether a hydrocarbon can be an open-chain (acyclic) compound is the general formula for saturated open-chain alkanes: \(C_nH_{2n+2}\). Any formula that matches this exactly can be drawn as a straight or branched chain with no ring. Formulae with fewer hydrogens than \(2n+2\) must contain either a double/triple bond or a ring (each ring or pi bond reduces the hydrogen count by 2).

For n = 6, the alkane formula is \(C_6H_{14}\), which corresponds to n-hexane and its branched isomers, all open-chain molecules with no ring.

The other three options have fewer hydrogens than \(C_6H_{14}\). \(C_6H_{10}\) fits cyclohexene (one ring plus one double bond), \(C_6H_6\) is benzene (a six-membered aromatic ring), and \(C_6H_8\) fits cyclohexadiene (a ring with two double bonds). Each of these must be drawn with a ring.

Hence, the molecular formula that does NOT represent a ring structure is \(C_6H_{14}\).

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Similar Questions

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  2. What do methane, ethane and propane represent in the study of carbon compounds?

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Important Questions from Hydrocarbons

  1. Which of the following base is found in soap?

  2. Which one of the following is the general formula of Alkenes?

  3. \({\rm{C}}{{\rm{H}}_3}{\rm{COOH}}\mathop \to \limits^{LiAI{H_4}} {\rm{A}}\mathop \to \limits^{PC{I_5}} {\rm{B}}\mathop \to \limits^{alc.KOH} {\rm{C}}\) Homologue of ‘C’ in the above reaction is:
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