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.
