Of the isomeric haexanes, the isomers that give the minimum and maximum number of monochloro derivatives are, respectively,

Options

A. `3`-methylpentane and `2,3`-dimethylbutane
B. `n`-hexane and `2,3`-dimethylbutane
C. `2,3`-dimethylbutane and `2`-methylpentane
D. `2,2`-dimethylbutane and `2`-methylpentane

Answer

C

Explanation

`n`-hexane gives `3` isomers, `2,3`-dimethylbutane gives `2`-isomers, and `2,2`-dimethylbutane gives `3` isomers:
`overset(1_(a)^(@))(CH)_(3)underset(2-"Methylpentane")(underset(1_(a)^(@)CH_(3))underset(|)overset(3^(@))(CH)overset(2_(a)^(@))(CH_(2))overset(2_(b)^(@))(CH_(2))overset(1_(b)^(@))CH_(3))` , `overset(1^(@))CH_(3)underset(2,3-"Dimethylbutane")(underset(1^(@)CH_(3))underset(|)overset(2^(@))(CH)underset(CH_(3))underset(|)overset(2^(@))(CH)underset(1^(@))overset(1^(@))(CH_(3))`