1. Principle involved: boiling point and molecular structure

   For hydrocarbons, boiling point mainly depends on:

   • molecular mass: higher mass generally increases boiling point,
   • shape / branching: more branching decreases surface area, so intermolecular van der Waals forces become weaker, hence boiling point decreases.

2. Compare the given compounds

   • A: $n$-butane = straight-chain alkane, $\mathrm{C_4H_{10}}$
   • B: isobutane = branched alkane, $\mathrm{C_4H_{10}}$
   • C: 1-butene = alkene, $\mathrm{C_4H_8}$
   • D: 1-butyne = alkyne, $\mathrm{C_4H_6}$

3. Key comparison between $n$-butane and isobutane

   Both have the same molecular formula $\mathrm{C_4H_{10}}$, but:

   • $n$-butane is straight chain
   • isobutane is branched

   Since branching reduces surface contact, isobutane has weaker intermolecular attraction and therefore a lower boiling point than $n$-butane.

4. Compare unsaturated compounds

   Although 1-butene and 1-butyne have fewer hydrogens, their boiling points are not lower than that of isobutane. Typical boiling points are approximately:

   • $n$-butane: about $-0.5^\circ \mathrm{C}$
   • isobutane: about $-11.7^\circ \mathrm{C}$
   • 1-butene: about $-6.3^\circ \mathrm{C}$
   • 1-butyne: about $8^\circ \mathrm{C}$

   Hence the minimum boiling point is for isobutane.

5. Option-wise conclusion

   • A: $n$-butane — not minimum
   • B: isobutane — minimum boiling point
   • C: 1-butene — higher than isobutane
   • D: 1-butyne — highest among these

Therefore, the correct answer is B.