Q is a terminal alkyne like 1-butyne Steps:

• Molecular formula C4H6 shows 2 degrees of unsaturation: expected hydrogens for alkane are 10, so (10-6)/2 = 2.
• Hydrogenation to single C4H8 product (1 degree of unsaturation) indicates partial reduction of triple bond to double bond, yielding 1-butene without isomers.
• This rules out dienes (multiple hydrogenation products) or cycloalkenes (different reactivity).
• Terminal alkyne structure fits, as internal alkynes like 2-butyne give symmetric but different Br2 products.

Why D is correct:

• Terminal alkynes react with Br2 (under substitution conditions) to replace acidic terminal H with Br, forming single bromoalkyne product C4H6Br per the formula given.

Why the others are wrong:

• A: Saturated hydrocarbons have formula C4H10; Q has C4H6, so unsaturated.
• B: Br2 addition to triple bond gives C4H6Br2 (dibromide), but multiple isomers possible, not single.
• C: Steam hydration of terminal alkyne follows Markovnikov's rule to give ketone C4H8O, not C4H6O.

Final answer: D