Question

Write the mechanism for the reaction involving [emim][Br] and lithium bis(trifluoromethylsulfonyl)imide.

Answer 1

It has a thermal decomposition mechanism followed as... the volatilization process for ionic liquids, the vapor evolved from heating the ionic liquid 1-ethyl-3-methylimidazolium bromide (EMIM(+)Br(-)) was analyzed via tunable vacuum ultraviolet photoionization time-of-flight mass spectrometry (VUV-PI-TOFMS) and thermogravimetric analysis mass spectrometry (TGA-MS). For this ionic liquid, the experimental results indicate that vaporization takes place via the evolution of alkyl bromides and alkylimidazoles, presumably through alkyl abstraction via an S(N)2 type mechanism, and that vaporization of intact ion pairs or the formation of carbenes is negligible. Activation enthalpies for the formation of the methyl and ethyl bromides were evaluated experimentally, ΔH(‡)(CH(3)Br) = 116.1 ± 6.6 kJ/mol and ΔH(‡)(CH(3)CH(2)Br) = 122.9 ± 7.2 kJ/mol, and the results are found to be in agreement with calculated values for the S(N)2 reactions. Comparisons of product photoionization efficiency (PIE) curves with literature data are in good agreement, and ab initio thermodynamics calculations are presented as further evidence for the proposed thermal decomposition mechanism. Estimates for the enthalpy of vaporization of EMIM(+)Br(-) and, by comparison, 1-butyl-3-methylimidazolium bromide (BMIM(+)Br(-)) from molecular dynamics calculations and their gas phase enthalpies of formation obtained by G4 calculations yield estimates for the ionic liquids' enthalpies of formation in the liquid phase: ΔH(vap)(298 K) (EMIM(+)Br(-)) = 168 ± 20 kJ/mol, ΔH(f, gas)(298 K) (EMIM(+)Br(-)) = 38.4 ± 10 kJ/mol, ΔH(f, liq)(298 K) (EMIM(+)Br(-)) = -130 ± 22 kJ/mol, ΔH(f, gas)(298 K) (BMIM(+)Br(-)) = -5.6 ± 10 kJ/mol, and ΔH(f, liq)(298 K) (BMIM(+)Br(-)) = -180 ± 20 kJ/mol.