When halogenated hydrocarbons strengthen alkaline KOH reagent in alcohol solvent, E2 type elimination reaction will occur. However, for E2-type elimination reaction to occur, the eliminated halogen atom and hydrogen atom (which can be any of the three isotopes H, D and T of hydrogen) must be * * * and 180 opposite in space. This is the special spatial specificity of E2 elimination reaction.

In the title of Fisher's projection of halogenated hydrocarbons, it is given that:

Because the single bond connecting carbon atom 1 and No.2 can rotate freely, part of Br atoms on carbon atom 1 and H atoms on carbon atom 2 of halogenated hydrocarbon molecules are in the * * * plane, which is opposite to the direction of 180, thus eliminating HBr molecules (as shown in Newman molecular conformation on the left in the figure below); However, there is still a part of halogenated hydrocarbon molecules, and the Br atom on carbon 1 and the D atom on carbon # 2 are * * * in opposite directions 180, which eliminates DBr molecules. (as shown in the Newman molecular conformation on the right in the figure below):