What happens in diffused sunlight?
Ethane undergoes free-radical substitution with excess $\mathrm{Br_2}$, so one or more H atoms can be replaced by Br.

Count all distinct brominated products of ethane
Ethane is $\mathrm{C_2H_6}$. Since the two carbon atoms are equivalent, distinct products arise only from different distributions of Br atoms on the two carbons.

Let the product be $\mathrm{C_2H_{6-n}Br_n}$, where $n=1$ to $6$.

We count distinct structural arrangements for each $n$:

For $n=1$: Monobromoethane

Only one product: $\mathrm{CH_3CH_2Br}$ Count $=1$

For $n=2$: Dibromoethanes

Possible distributions of 2 Br on the two carbons:

• $(2,0)$: $\mathrm{CH_3CHBr_2}$ (1,1-dibromoethane)
• $(1,1)$: $\mathrm{CH_2BrCH_2Br}$ (1,2-dibromoethane) Count $=2$

For $n=3$: Tribromoethanes

Possible distributions:

• $(3,0)$: $\mathrm{CH_3CBr_3}$ (1,1,1-tribromoethane)
• $(2,1)$: $\mathrm{CH_2BrCHBr_2}$ (1,1,2-tribromoethane) Count $=2$

For $n=4$: Tetrabromoethanes

Possible distributions:

• $(3,1)$: $\mathrm{CH_2BrCBr_3}$ (1,1,1,2-tetrabromoethane)
• $(2,2)$: $\mathrm{CHBr_2CHBr_2}$ (1,1,2,2-tetrabromoethane) Count $=2$

For $n=5$: Pentabromoethane

Possible distribution:

• $(3,2)$: $\mathrm{CHBr_2CBr_3}$ Count $=1$

For $n=6$: Hexabromoethane

Only one product: $\mathrm{CBr_3CBr_3}$ Count $=1$

Total number of bromo derivatives $1+2+2+2+1+1 = 9$

Conclusion Therefore, the number of bromo derivatives formed is: $\boxed{9}$

Comparison with stored answer
Stored correct answer $=9$, which matches the derived result.