Science (086) — AI-generated practice question

Structural Isomers of Butane (C₄H₁₀):

1. n-Butane (Normal butane) — straight chain

$$\text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{CH}_3$$

2. Isobutane (2-methylpropane) — branched chain

$$\text{CH}_3 - \overset{|}{\underset{|}{\text{CH}}} (\text{CH}_3) - \text{CH}_3$$

$$\text{i.e., } \text{CH}_3{-}\text{CH(CH}_3){-}\text{CH}_3$$

Differing Physical Property — Boiling Point:
n-Butane has a higher boiling point (−1°C) than isobutane (−12°C).

Reason: n-Butane has a longer, straight chain, allowing greater surface area for intermolecular interactions. Isobutane's branched, compact shape reduces surface contact, resulting in weaker intermolecular forces and hence a lower boiling point.

Source: Carbon and its Compounds, Section 4.2.2

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• 2 marks are for drawing both correct structural formulae with common names — examiners check that all C and H valencies are satisfied and the chain/branch is clearly shown.
• 1 mark is for naming a physical property and giving a reason. Boiling point is the safest choice; melting point is also acceptable.
• Write the branched structure clearly, showing the side —CH₃ group attached to the middle carbon; don't just write the molecular formula.
• The reason must link molecular shape → surface area → intermolecular forces → boiling point to earn the mark; just saying "different structures" is not enough.