The reaction you provided involves the use of LDA (Lithium Diisopropylamide) and 1-bromopropane (CH₃CH₂CH₂Br). LDA is a strong base commonly used in organic synthesis. The reaction you described is likely an example of an elimination reaction, where LDA acts as a base to remove a proton (deprotonate) from the 1-bromopropane molecule, resulting in the formation of an alkene.
The predicted product of the reaction would be 1-butene (CH₃CH₂CH=CH₂). LDA removes a proton (H⁺) from the β-carbon adjacent to the bromine atom, leading to the formation of a double bond (alkene) between the β- and γ-carbons.
Please note that the reaction conditions, such as temperature and solvent, can influence the reaction outcome. Without specific details about the reaction conditions, it is difficult to provide a precise prediction.