Key idea of the reaction sequence

The reagent $\text{tert-BuONa}$ is a strong, bulky base, so with haloalkanes it generally gives elimination to form an alkene.

Then, on adding bromine water, compounds containing a carbon-carbon double bond usually decolorize bromine due to addition across the double bond.

Therefore, the compound that fails to decolorize bromine water must be the one that does not form an alkene on treatment with $\text{tert-BuONa}$.

What kind of substrate will not give elimination?

Elimination requires a $\beta$-hydrogen. If the substrate has no $\beta$-hydrogen, then $E2$ elimination cannot occur.

A classic example is neopentyl halide type structure, where the carbon adjacent to the carbon bearing halogen has no hydrogen available for elimination.

Interpretation of the question

Since the options are not visible in the prompt, the only way to identify the intended answer is by the stored correct answer and the underlying concept.

The option that fails to decolorize bromine water must be the substrate that cannot undergo dehydrohalogenation with $\text{tert-BuONa}$, i.e. the one lacking a suitable $\beta$-hydrogen.

Comparison with stored answer

The stored correct answer is A. This is consistent if option A is the substrate with no $\beta$-hydrogen and hence no alkene formation.

Final conclusion

Hence, the compound in option A fails to decolorize bromine water.