Question

Hydroboration and Oxidation Experiment

We reacted ammonia borane (4 mmol) and 1-octene (12 mmol).

After that, we added 1400 microliter of 3M NaOH solution and 1400 microliter of 30 % H2O2.

How much excess NaOH and H2O2 were used in this experiment?

Answer 1

Volume of NaOH used = 1400 uL = 1.4 * 10^(-3) L

Number of moles of NaOH = Volume of solution (in L) * Molarity of NaOH (M)

=> 1.4 * 10^(-3) * 3

=> 4.2 * 10^(-3) moles

Molar mass of NaOH = 23 + 16 +1 = 40 gm/mol

Mass of NaOH used = number of moles * molar mass = 4.2 * 10^(-3) mol * 40 gm/mol = 168 * 10^(-3) gm = 0.168g

Volume of H2O2 added = 1400 uL = 1.4 * 10^(-3) L

Assuming it is 30% (w/v) solution. 1.4 * 10^(-3) L * 30 gm/(100 mL) * (1000 mL/1L) = 1.4 * 10^(-3) * 300 = 420 * 10^(-3) = 0.42g

Hence the mass of H2O2 added the solution = 0.42g