I'm not sure if this is correct. Googling fexprs takes me to this Lambda the Ultimate post:
http://lambda-the-ultimate.org/node/3640#comment-51665. There it says that fexprs map from s-expressions (of the operands to fexpr) to a value (which replaces the fexpr call). L2's macro system does not do this. L2 maps from s-expressions (of the operands to the macro) to s-expressions (that are supposed to replace the macro call).