Question

Which of the following alkanes exists in liquids state at room temperature?
A.\[{\text{C}}{{\text{H}}_{\text{4}}}\]
B.\[{{\text{C}}_{\text{2}}}{{\text{H}}_6}\]
C.\[{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}\]
D.\[{{\text{C}}_{\text{5}}}{{\text{H}}_{12}}\]

Answer 1

Hint: Alkanes or paraffin are saturated hydrocarbons. They are very less reactive in comparison to the other organic compounds. The lower hydrocarbons have weak Vander Waal forces and are present in gaseous state.

Complete step by step solution:
A. \[{\text{C}}{{\text{H}}_{\text{4}}}\] - methane is the first member of the alkane family. It is also known by the name “marsh gas” and exists at room temperature in the gaseous state.
B. \[{{\text{C}}_{\text{2}}}{{\text{H}}_6}\] - ethane is the second member of the alkane family and it also exists as gas at room temperature.
C. \[{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}\] - butane is the fourth member of the alkane family and it exists in two isomeric forms: one as n-butane and the other as but-2-ene. Both the isomers exist in the gaseous form.
D. \[{{\text{C}}_{\text{5}}}{{\text{H}}_{12}}\] - pentane is the fifth member of the alkane family and it also exists as two isomers, n-pentane and 2- pentane. This exists as liquid at room temperature.

Hence, option D is correct.

Note:
Branched alkanes normally exhibit lower boiling points than unbranched alkanes of the same carbon content. This occurs because of the greater van der Waals force that exists between molecules of the unbranched alkanes. These forces can be dipole-dipole, dipole-induced dipole, or induced dipole-induced dipole in nature. The unbranched alkanes have greater force of attraction because of their greater surface area.
Solid alkanes are normally soft with lower melting points. These characteristics are due to the strong repulsive forces generated between electrons on the neighbouring atoms which are in close proximity in crystalline solids.