Primary amines — practice questions

Step 1: Formation of the nitrile React bromoethane with potassium cyanide (KCN) in ethanol. This is a nucleophilic substitution reaction. Equation: $CH_3CH_2Br + KCN \xrightarrow{ethanol, reflux} CH_3CH_2CN + KBr$ Product: Propanenitrile.

Step 2: Reduction of the nitrile Reduce the propanenitrile to form propylamine. Use a suitable reducing agent like $LiAlH_4$ in dry ether. Equation: $CH_3CH_2CN + 4[H] \xrightarrow{1. LiAlH_4 \text{ in dry ether}} \xrightarrow{2. H_2O} CH_3CH_2CH_2NH_2$ Final Product: Propylamine.

Step 1: Write the balanced equation. $CH_3CH_2CH_2CH_2NH_2 + CH_3COCl \rightarrow CH_3CONHCH_2CH_2CH_2CH_3 + HCl$

Step 2: Calculate the molar mass of butan-1-amine. Mr(butan-1-amine) = (4 × 12.0) + (11 × 1.0) + 14.0 = 48.0 + 11.0 + 14.0 = 73.0 g mol^{-1}

Step 3: Calculate the moles of butan-1-amine. Moles = mass / Mr = 1.20 g / 73.0 g mol^{-1} = 0.01644 mol

Step 4: Use stoichiometry. The molar ratio of butan-1-amine to the amide product (N-butylethanamide) is 1:1. Therefore, moles of amide = 0.01644 mol.

Step 5: Calculate the molar mass of the amide product. Product is N-butylethanamide, $CH_3CONH(CH_2)_3CH_3$. Formula: $C_6H_{13}NO$ Mr(amide) = (6 × 12.0) + (13 × 1.0) + 14.0 + 16.0 = 72.0 + 13.0 + 14.0 + 16.0 = 115.0 g mol^{-1}

Step 6: Calculate the mass of the amide product. Mass = moles × Mr = 0.01644 mol × 115.0 g mol^{-1} = 1.89 g (to 3 s.f.)

Answer: 1.89 g