Answer: D (the same homologous series as CH₄). The compound C₄H₁₀ fits the alkane general formula \(\text{C}_n\text{H}_{2n+2}\) with n = 4, so it is a saturated alkane (butane). Methane, CH₄, is the first member of that same alkane homologous series, so statement D is correct.
The other statements are wrong. C₄H₁₀ is saturated, so it does not decolourise bromine water — only unsaturated alkenes do that, making A wrong. Boiling point increases with chain length because larger molecules have stronger intermolecular forces, so C₄H₁₀ (butane) has a higher boiling point than C₃H₈ (propane), not a lower one, so B is wrong.
C₄H₁₀ has only two isomers — butane (a straight chain) and 2-methylpropane (one branched chain) — so it has just one branched-chain isomer, not two, making C wrong. Only D is correct.
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