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  1. Problems
  2. /A chemist treats 5.45 g of bromoethane (109 g/mol) with alcoholic potassium cyanide to make…
EasyChemistryHigh School (11–12)Organic Compounds with Functional Groups › AminesNumerical

A chemist treats 5.45 g of bromoethane (109 g/mol) with alcoholic potassium cyanide to make…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A chemist treats 5.45 g5.45\ \text{g}5.45 g of bromoethane (109 g/mol109\ \text{g/mol}109 g/mol) with alcoholic potassium cyanide to make propanenitrile, C3H5N\text{C}_3\text{H}_5\text{N}C3​H5​N (55 g/mol55\ \text{g/mol}55 g/mol). All of the halide reacts. The nitrile is then either boiled with dilute acid, to give propanoic acid (74 g/mol74\ \text{g/mol}74 g/mol) and ammonia, or reduced to propan-1-amine (59 g/mol59\ \text{g/mol}59 g/mol). Take every step at a yield of 100 %. Find the mass of propanenitrile, the mass of propanoic acid, the mass of ammonia formed in the hydrolysis (17 g/mol17\ \text{g/mol}17 g/mol) and the mass of propan-1-amine.

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Need a reminder? The key ideaShowHide

Silver cyanide gives an isocyanide and potassium cyanide gives a nitrile. A nitrile is reduced to a primary amine with one more carbon than the halide. An isocyanide is hydrolysed by acid to a primary amine and formic acid, and is reduced to a secondary amine. Nitrobenzene takes 6, 4 or 5 electrons in an acidic, neutral or alkaline medium, and each atom of zinc or tin gives 2 electrons.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Amount. 5.45109=0.0500 mol\dfrac{5.45}{109} = 0.0500\ \text{mol}1095.45​=0.0500 mol.

Step 2. Nitrile. 0.0500×55=2.75 g0.0500 \times 55 = 2.75\ \text{g}0.0500×55=2.75 g.

Step 3. Acid. 0.0500×74=3.70 g0.0500 \times 74 = 3.70\ \text{g}0.0500×74=3.70 g. The ammonia is 0.0500×17=0.850 g0.0500 \times 17 = 0.850\ \text{g}0.0500×17=0.850 g.

Step 4. Amine. 0.0500×59=2.95 g0.0500 \times 59 = 2.95\ \text{g}0.0500×59=2.95 g.

Answer: 2.75 g2.75\ \text{g}2.75 g, 3.70 g3.70\ \text{g}3.70 g, 0.850 g0.850\ \text{g}0.850 g and 2.95 g2.95\ \text{g}2.95 g.

Check. The nitrile adds one carbon to the two carbons of bromoethane, so the acid and the amine both have three carbon atoms.

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  1. Cyanides, isocyanides and nitro compoundsDiazonium Salts and Other Nitrogen Compounds

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