Romanian boxer Lucian Bute will defend his IBF super middleweight world title in a bout against France’s Jean-Paul Mendy on July 9 in Bucharest, realitatea.net informs. Jean-Paul Mendy (38), the mandatory title challenger, is undefeated. Mendy won 29 bouts, 16 through KO, and plans to unify all super middleweight titles provided he defeats Lucian Bute.
The Romanian boxer will start training in May in Montreal (Canada) and will then leave for Milan (Italy) in June. Bute will arrive in Romania on July 2, a week before the bout.
According to Rudel Obreja, President of the Romanian Boxing Federation, the place where the bout will take place is unknown at this point, with the details set to be arranged in the following days. The Romexpo centre and the Polivalenta Hall are the most likely locations.
“Nothing has been established for the time being in what concerns the location of the bout, everything will be worked out in the next two or three days.
The possible locations are: Romexpo centre, Polivalenta Hall and a stadium. The first two are top contenders. Let me tell you how things are: on one hand we have a higher rent and on the other a higher TV broadcasting potential. The ideal solution would be a combination between the two,” Rudel Obreja stated for HotNews.ro.
Lucian Bute is undefeated in 28 professional bouts, winning 23 of them through KO. The Romanian won the title in October 2007, defeating Columbia’s Alejandro Berrio through KO in round 11. Until now Bute successfully defend his title seven times: in February 2008 against America’s William Joppy (KO), in October 2008 against Mexico’s Libdrado Andrade (points), in March 2009 against Columbia’s Fulgencio Zuniga (technical KO), in November 2009 against Mexico’s Librado Andrade (KO), in April 2010 against Columbia’s Edison Miranda (KO), in October 2010 against America’s Jesse Brinkley (technical KO) and in March 2011 against Ireland’s Brian Magee.