Question

ental data shown below to determine the actual (empirical) rate law for the Use the experim reaction between trityl bromide and iodide. Compare this experimental the Sx1 and Sy through an SN1 (two-step) or Sy2 (one-step) mechanism 4. ly-determined rate law to the predicted rate laws you determined for 2 mechanisms, in the previous problem, and determine if this reaction proceeds (Cis8H1s)C-Br (M) 0.035 0.105 0.105 Experiment Rate (M/s) 0.020 0.020 0.060 0.060 1.2x10-6 3.6x10-6 3.6x106 4.8x10-5 2 4 0.140 (a) What is the experimental rate law for the reaction between trityl bromide and iodide? (b) Does this reaction go through an S,2 (single step) or Syl (two-step) mechanism? Explain your answer

Answer 1

Comparing experiment Number 1 and 2 here rate of I^- is constant and when we triple the concentration of (C_18 H_15) C - Br, Rate also become triple that is order with respect to (c_18 H_1s) C - Br is first order Now dividing experiment Number 1 6y (0.035/0.140)^A (0.02/0.06)^m = 1.2 times 10^-6/4.8 times 10^-6 here n and m are the order of (C_18 H_1s) C - Br and I^- respectively as we know n = 1 (0.035/0.140)^1 (1/3)^m = 1.2/4.8 rightarrow (1/3)^m = 1.2 times 0.14/0.035 times 4.8 From equation number (1) (1/3)^m = (1/1)^1 now taking ln on both sides m(ln 1 - ln3) = 0 -m ln3 = 0 From equation number = 3 m times 1.09 = 0 or 1 times m = 0 that order with respect to I^- is zero order So for this reaction, overall rate law = [(C_18 H_15) C - Br]^0 [I^-]^1 as the order of reaction = 0[with respect to C_18 H_15) C - Br] + 1 [with respect to I^-]