CHEM2133 QUIZ 2 – 50 points 1. Preparation of t‐BuNH2BH3 i. (5 points) Balance the reaction below: (no partial credit)
ii.
(5 points) You perform the reaction in the hood. There must be some hazardous chemicals form both starting materials and products. State their names or draw their structures t-butylamine is highly toxic; highly flammable; corrosive. Hydrogen is highly flammable. Boranetert-butylamine complex is harmful to skin and very toxic. Ammonia.
iii.
(5 points) Percentage yield would decrease if tetrahydrofuran is used as a solvent instead of toluene. Why?
iv.
THF coordinate with BH3 (5 points) Before removing the reaction flask from the hood, you needed to heat the flask for 10 minutes. State the reason why you had to do that. To remove excess t-butylamine and ammonia gas.
v.
(5 points) What is petroleum ether? c. A mixture of alkanes 2. Preparation of t‐BuNH2BH2I vi. (5 points) Balance the reaction below: (no partial credit)
vii.
(5 points) When you added I2 to react with tert‐butylamine borane, you didn’t want to add excess I2. Explain how you would do this step. Add small amount at a time until it slightly turns brown, stop adding I2 and add small amount of tbutylamine borane until the solution becomes colorless.
3. Reactivity comparison of magnesium and aluminum vii. (5 points) How would it affect the volume of H2 gas produced from reaction of Mg with hydrochloric acid if you used concentrated hydrochloric acid (35%) instead of 2% hydrochloric acid? Number of moles of H2 gas will not be affected but the rate of gas producing will be higher so reaction will reach the completion faster.
viii.
(10 points) Calculate theoretical volume of H2 produced from reaction of 10.9 g of Al and excess sodium hydroxide solution. Write and balance the chemical reaction for full credit. (Al = 27, H = 1.008) Gas Law: PV = nRT; P = 1 atm, n = mol of gas, R = 0.0821 L∙atm∙K‐1∙mol‐1, T = temperature 2Al (s) + 2NaOH (aq) + 6H2O (l) 2NaAl(OH)4 (aq) + 3H2 (g) 2 mol of Al will produce 3 mol of H2 gas. 10.9 g of Al = 0.404 mol of Al 0.606 mol of H2 gas Gas Law: PV = nRT; P = 1 atm, n = mol of H2, R = 0.0821 L·atm·K-1·mol-1, T = temperature in Kelvin (1 atm)(V) = (0.606 mol)(0.0821 L·atm·K-1·mol-1)(298 K) V = 14.8 L