1bromo2methylpropane 2li Diethyl Ether
The reaction of halogenated hydrocarbons with lithium in ether
Now there is 1-bromo-2-methylpropane. Take an appropriate amount of this halogenated hydrocarbon, place it in a clean reaction vessel, add a sufficient amount of lithium (Li), and use anhydrous Diethyl Ether as a solvent.
Under suitable reaction conditions, the bromine atom (Br) in halogenated hydrocarbons has high activity, and the outer layer of lithium atom (Li) has only one electron. It is chemically active and easy to lose electrons. The lithium atom will react with the bromine atom in 1-bromo-2-methylpropane. The lithium atom loses electrons and becomes a lithium ion (Li 🥰). The bromine atom gains electrons to form a bromine ion (Br), and then generates an organolithium reagent and lithium bromide (LiBr). In this reaction process, Diethyl Ether acts as a solvent to dissolve the reactants, so that the reaction can proceed smoothly in the homogeneous system, promoting the contact and collision between the reaction molecules, which is conducive to the occurrence of the reaction.
This reaction is of great significance in the field of organic synthesis. The generated organolithium reagent can be used as a powerful reagent in many organic reactions such as the construction of carbon-carbon bonds, providing an important way and method for the synthesis of organic compounds.
