Worked example 1

Propanoyl chloride ( $CH_3CH_2COCl$ ) is reacted with propan-2-ol. (i) Name and draw the structure of the organic product formed. (ii) Write a balanced chemical equation for the reaction.

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(i) Product Name: Isopropyl propanoate (or 1-methylethyl propanoate).

Structure: The acyl group from propanoyl chloride ( $CH_3CH_2CO-$ ) attaches to the oxygen of propan-2-ol, replacing the H. $CH_3CH_2COOCH(CH_3)_2$

(ii) Equation: $CH_3CH_2COCl + (CH_3)_2CHOH \rightarrow CH_3CH_2COOCH(CH_3)_2 + HCl$

Marking points: Correct ester structure [1], correct name [1], correct balanced equation showing HCl by-product [1].

Worked example 2

Calculate the mass of N-ethylethanamide ( $CH_3CONHCH_2CH_3$ ) formed when 5.00 g of ethanoyl chloride ( $CH_3COCl$ ) is added to an excess of ethylamine ( $CH_3CH_2NH_2$ ). ( $A_r$ values: C=12.0, H=1.0, O=16.0, N=14.0, Cl=35.5)

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Step 1: Write the balanced equation. $CH_3COCl + 2CH_3CH_2NH_2 \rightarrow CH_3CONHCH_2CH_3 + CH_3CH_2NH_3Cl$

Step 2: Calculate the molar masses. $M_r(CH_3COCl) = (2 \times 12.0) + (3 \times 1.0) + 16.0 + 35.5 = 78.5\ g\ mol^{-1}$ $M_r(CH_3CONHCH_2CH_3) = (4 \times 12.0) + (9 \times 1.0) + 16.0 + 14.0 = 87.0\ g\ mol^{-1}$

Step 3: Calculate the moles of the limiting reactant (ethanoyl chloride). Moles of $CH_3COCl = \frac{mass}{M_r} = \frac{5.00\ g}{78.5\ g\ mol^{-1}} = 0.06369\ mol$

Step 4: Use the stoichiometry to find the moles of the product. The molar ratio of $CH_3COCl$ to $CH_3CONHCH_2CH_3$ is 1:1. Therefore, moles of N-ethylethanamide = 0.06369 mol.

Step 5: Calculate the mass of the product. Mass = moles $\times M_r = 0.06369\ mol \times 87.0\ g\ mol^{-1} = 5.54\ g$ (to 3 s.f.)

Answer: 5.54 g

Acyl chlorides — practice questions

Worked example 1

Worked example 2