Match the boiling points (°ree;C, 1 atm) with the appropriate alkanes. Be sure to answer all parts. A) 2,2−dimethylbutane B) heptane C) hexane D) 2−methylpentane

Explanation:

A) 2,2−dimethylbutane - 79°C

B) heptane - 126°C

C) hexane - 98°C

D) 2−methylpentane - 90°C

As the number of C-C chain increases in the homologous series(alkane), the van der waals dispersion forces increases because the atoms increase i.e Heptane (C7) has a stronger van der waal force than Hexane (C6).

As branching increases (more branches) volatility also increases and temperature is a factor of volatility. As the volatility increases, the boiling point decreases.

2,2−dimethylbutane has more branches (2) than 2−methylpentane, so 2,2−dimethylbutane is more volatile than 2−methylpentane so hence 2,2−dimethylbutane is lesser.

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