Constitutional isomers in free radical chlorination of ethane

Steps:

• List isomers by substitution level: 1 mono (C₂H₅Cl), 2 di (1,1- and 1,2-), 3 tri (1,1,1-; 1,1,2-; 1,2,2-), 2 tetra (1,1,1,2- and 1,1,2,2-), 1 penta (1,1,1,2,2-), 1 hexa (1,1,1,1,2,2-).
• Total isomers: 1+2+3+2+1+1=10.
• Products with two or more Cl atoms exclude the mono isomer.
• Count for n≥2: 2+3+2+1+1=9.

Why D is correct:

• Free radical chlorination with excess Cl₂ yields all constitutional isomers of C₂H_{6-n}Cl_n (n=1–6), totaling 9 with n≥2 due to ethane's symmetry and possible substitutions.

Why the others are wrong:

• A (6): Ignores higher substitutions like tetra, penta, hexa isomers.
• B (7): Excludes penta- and hexachloroethane.
• C (8): Omits hexachloroethane (C₂Cl₆).

Final answer: D