Isopentane is allowed to undergo free-radical chlorination and the reaction mixture is separated. How many optically active fractions of formula `C_(5)H_(11)CI` would you expect to collect?

Options

A. Zero
B. Two
C. One
D. Four

Answer

A

Explanation

Four fractions are collected but no fraction is optically active because each fraction is eithe a single achiral compound or a racemic modification (Inactive reactants give inactive products)
`overset(CH_(3))overset(|)underset("Isopetane")(CH_(3)CH_(2)CHCH_(3))overset(CI_(2))rarroverset(CH_(3))overset(|)underset(1-"Chloro"-2-"methy1 butane")underset(**)(CH_(3)CH_(2)CHCH_(2)CI) overset(CH_(3))overset(|)underset(2-"Chloro-2-methy1 butane")underset(CI)underset(|)(+CH_(3)CH_(2)C CH_(3))+underset(2-"Chloro-3-methy1 butane")underset(CI)underset(|)(CH_(3)overset(**)CHCHoverset(CH_(3))overset(|)(CHCH_(3)))overset(CH_(3))overset(|)underset(1-"Chloro-3-methy1 butane")(+CICH_(2)CH_(2)CHCH_(3))`
Fraction 1

Fraction 2

Fraction 3, Fraction 4
`overset(CH_(3))overset(|)underset(2-"Chloro-2-methylbutane")underset(CI)underset(|)(CH_(3)H_(2)C-CH_(3)) overset(CH_(3))overset(|)underset(1-"Chloro-3-methylbutane")underset(H)underset(|)(CICH_(2)CH_(2)-C-CH_(3))`
Replacement of eithe `H` on`-CH_(2)-`(in isopentane) to give `2-`chloro`-3-`methylbutane is equally likely and replacement of any `H` on either `-CH_(3)`(in isopentan) to given `1-`chloro`2-`methylbutane is equally likely, in each case giving rise to the racemic modification.