Amides — practice questions

(i) The product is N-ethylbutanamide. The butanoyl part provides the four-carbon chain with the carbonyl group, and the ethylamine provides the ethyl group attached to the nitrogen. \ Structure: CH₃-CH₂-CH₂-C(=O)-NH-CH₂-CH₃ \ \ (ii) Two moles of ethylamine are required. One acts as the nucleophile to attack the carbonyl carbon of butanoyl chloride, and the second acts as a base to accept the proton from the intermediate and neutralise the HCl byproduct. \ Equation: \ CH₃CH₂CH₂COCl + 2CH₃CH₂NH₂ → CH₃CH₂CH₂CONHCH₂CH₃ + CH₃CH₂NH₃⁺Cl⁻

Step 1: Write the balanced equation for the reaction. \ CH₃CH₂CONH₂ + H₂O + H⁺ → CH₃CH₂COOH + NH₄⁺ \ \ Step 2: Calculate the molar mass of propanamide. \ Mr(CH₃CH₂CONH₂) = (3 × 12.0) + (7 × 1.0) + 14.0 + 16.0 = 73.0 g mol⁻¹ \ \ Step 3: Calculate the moles of propanamide used. \ Moles = mass / Mr = 5.85 g / 73.0 g mol⁻¹ = 0.08013... mol \ \ Step 4: Use the stoichiometry of the reaction to find the moles of propanoic acid produced. \ The molar ratio of propanamide to propanoic acid is 1:1. \ Therefore, moles of propanoic acid = 0.08013... mol. \ \ Step 5: Calculate the mass of propanoic acid. \ Mr(CH₃CH₂COOH) = (3 × 12.0) + (6 × 1.0) + (2 × 16.0) = 74.0 g mol⁻¹ \ Mass = moles × Mr = 0.08013... mol × 74.0 g mol⁻¹ = 5.929... g \ \ Final Answer (to 3 s.f.): 5.93 g