Answer :
If water were added to an alkyne and the product was 2-pentanone, then the alkyne must have been 1-pentyne. 2-pentyne would have given a mixture of 2- and 3-pentanone.
The enol that would give 2-pentanone would have been pent-1-en-2-ol. Because an equilibrium favors the ketone so greatly, equilibrium is not a good description. If the ketone were treated with bromine, little reaction would be seen as the enol content would be too low. If a catalyst were added, NaOH for example, then formation of the enolate of pent-1-en-2-ol would form and react with bromine. This would eventually give a bromoform product. Under acidic conditions, the enol would favor formation of the more substituted enol consistent with alkene stability.
The enol that would give 2-pentanone would have been pent-1-en-2-ol. Because an equilibrium favors the ketone so greatly, equilibrium is not a good description. If the ketone were treated with bromine, little reaction would be seen as the enol content would be too low. If a catalyst were added, NaOH for example, then formation of the enolate of pent-1-en-2-ol would form and react with bromine. This would eventually give a bromoform product. Under acidic conditions, the enol would favor formation of the more substituted enol consistent with alkene stability.