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The “naked” fluoride [NMe4]F in CH2Cl2 converts the salt-like compound [C6F5Xe][AsF6] into the asymmetric, hypervalent molecule 1 in 70 % yield. As a result of its high fluoride-donating ability and its permanent dipole moment, 1 reacts in good yields with Cd(C6F5)2 to form Xe(C6F5)2 (2) and with Me3SiCN to form C6F5XeCN (3).
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Frohn et al. (2000) studied this question.
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