Dear Vida Ravankhah, the synthesis of Na[BH(mim)3] is quite similar to the synthesis of Trofimenkos famous Potassium-hydrotris(pyrazoly)borates ("Scorpionates"). It is a melt reaction that requires no solvent! You can follow the prodeeding of the reaction by measuring the amount of hydrogen gas evolved. The elimination of hydrogen gas requires a high temperature. The melt should be clear white, but a light brown color is als possible. Good luck!

Dear Vida Ravankhah, NaBH4 reacts with 3 equiv. of 2-methylimidazole (= 2-mimH) according to the following equation:

NaBH4 + 3 2-mimH = K[HB(2-mim)3] + 3 H2

Thus 3 protons of the 2-methylimidazole combine with 3 hydride ions of the NaBH4 to give 3 equiv. of elemental hydrogen. The structure of the product, K[HB(2-mim)3], is depicted in the attached formula drawing. In order to finally replace three hydride ions and get the tri-substituted product, the reaction must be done in a melt at 200 °C. The starting materials are mixed as solids (preferably under nitrogen!) and then heated to 200 °C until no more hydrogen is evolved. Then the work-up is done as described. Tomorrow I can send you a diagram of how the apparatus could look like.

Dear Vida Ravankhah, as promised, please find attached a schematic representation of the experimental set-up for your synthesis. It's just drawn by hand, but should be fairly clear. I hope you can find all the parts in your lab. Important is that the melt of the solid starting materials is heated slowly to 200 °C in an oil bath with constant stirring. With the measuring cylinder you can check if the calculated amount of hydrogen (3 equiv.) is evolved. If you have any further questions about this, please don't hesitate to ask me.