Bromine test distinguishes saturated from unsaturated hydrocarbons Steps:

• Bromine decolorizes in the dark via electrophilic addition to pi bonds in alkenes/alkynes.
• No decolorization means no pi bonds, so Q is saturated: acyclic alkane (C_nH_{2n+2}) or cycloalkane (C_nH_{2n}).
• For n=4, acyclic is C4H10; cyclo is C4H8; both have structural isomers and fit the test.
• C4H6 (C_nH_{2n-2}) always has pi bonds (double/triple or equivalent), so decolorizes.

Why B is correct:

• C4H8 fits if Q is cycloalkane isomer.

Why the others are wrong:

• A: C4H10 fits equally well as acyclic alkane (no pi bonds).
• C: C4H6 has pi bonds in all isomers, so decolorizes.
• D: Same as A; fits but doesn't distinguish from B.

Not enough information. Final answer: Not enough information.