Question

Methane, $CH_{4}$, reacts with chlorine in the presence of light gives four products: methyl chloride ($CH_{3}Cl$), dichloro methane ($CH_{2}Cl_{2}$), trichloromethane ($CHCl_{3}$) and tetrachloro-methane ($CCl_{4}$). Which compound has the lowest vapour pressure at room temperature? Explain.

Answer 1

    The reaction of Methane  $CH_{4}$  with chlorine under sunlight or UV light or at $400^{ \circ }C$ . “Hydrogen atoms in the Methane” are replaced by the chlorine atoms and give the following four products:

Methyl chloride $CH_{3}Cl$,

Dichloro methane $CH_{2}Cl_{2}$,

Trichloromethane $CHCl_{3}$,

Tetrachloro-methane $CCl_{4}$.  

Out of all these products, Tetrachloromethane has the lowest vapour pressure at the room temperature. It is because of the fact that Tetrachloromethane the molecular mass and the greatest London forces are high.

Answer 2

Explanation:

It is a substitution reaction where one hydrogen atom of alkane is substituted by chlorine atom and it forms an alkyl chloride and this reaction takes place only in the presence of light and an alkyl halide is formed alongwith hydrogen chloride.