Propane undergoes free-radical monochlorination at 300 K. Take the relative reactivity per hydrogen as primary : secondary $= 1 : 3.8$ . The percentage of 2-chloropropane in the monochloro product mixture is nearest to

Answer: B

Propane has 6 primary and 2 secondary hydrogens. Weighted yields: primary

6 \times 1 = 6

, secondary

2 \times 3.8 = 7.6

. Fraction of 2-chloropropane

= 7.6/13.6 = 0.559

, i.e. about 55.9%.