Carboxylic Acids and Derivatives: Question 4

Worked solution

Part (a): Preparing propanoyl chloride

Carboxylic acids are converted into the much more reactive acyl chlorides using sulfur dichloride oxide (thionyl chloride), $\text{SOCl}_2$, phosphorus(III) chloride, $\text{PCl}_3$, or phosphorus(V) chloride, $\text{PCl}_5$. With $\text{SOCl}_2$:

$\text{CH}_3\text{CH}_2\text{COOH} + \text{SOCl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{COCl} + \text{SO}_2 + \text{HCl}$

Word equation: propanoic acid + sulfur dichloride oxide $\rightarrow$ propanoyl chloride + sulfur dioxide + hydrogen chloride.

Advantage over $\text{PCl}_5$: both by-products of the $\text{SOCl}_2$ reaction, sulfur dioxide and hydrogen chloride, are gases that simply escape from the reaction mixture, leaving the propanoyl chloride behind with no need for further separation. Using $\text{PCl}_5$ instead also produces phosphorus oxychloride, $\text{POCl}_3$, a liquid by-product that has to be separated from the acyl chloride, making purification more difficult.

Part (b): Hydrolysis with water

Acyl chlorides are far more reactive than the parent carboxylic acid towards nucleophiles because chlorine is a much better leaving group than the hydroxide that would have to leave from an acid, and the electronegative chlorine also makes the carbonyl carbon strongly electrophilic. Propanoyl chloride therefore reacts rapidly with cold water by addition-elimination (hydrolysis):

$\text{CH}_3\text{CH}_2\text{COCl} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{COOH} + \text{HCl}$

Observation: a vigorous reaction accompanied by steamy white fumes of hydrogen chloride gas. Products: propanoic acid and hydrogen chloride.

Part (c): Reaction with an alcohol

Adding propanoyl chloride dropwise to excess ethanol again proceeds by addition-elimination, forming an ester directly (no catalyst or reflux needed, unlike direct esterification of the acid):

$\text{CH}_3\text{CH}_2\text{COCl} + \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3 + \text{HCl}$

Organic product: ethyl propanoate. Observation: steamy white fumes of hydrogen chloride are given off as the reaction proceeds.

Part (d): Reaction with an amine

Propanoyl chloride reacts with methylamine, again by addition-elimination, to form a substituted (secondary) amide:

$\text{CH}_3\text{CH}_2\text{COCl} + \text{2CH}_3\text{NH}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CONHCH}_3 + \text{CH}_3\text{NH}_3\text{Cl}$

Main organic product: N-methylpropanamide. Why excess methylamine is used: the first mole of methylamine reacts with the acyl chloride to give the amide and HCl. Because methylamine is a base, if no further amine were present the HCl produced would react with (protonate) methylamine or even the amide product, reducing the yield. Using an excess means the second mole of methylamine instead reacts with this HCl, forming the salt methylammonium chloride, $\text{CH}_3\text{NH}_3\text{Cl}$, and leaving the amide product intact.

Final answers

• (a) Propanoic acid + $\text{SOCl}_2$ $\rightarrow$ propanoyl chloride + $\text{SO}_2$ + HCl; both by-products are gases, so no separation step is needed (unlike with $\text{PCl}_5$).
• (b) Vigorous reaction, steamy white fumes of HCl; products are propanoic acid and hydrogen chloride.
• (c) Ethyl propanoate formed; steamy white fumes of HCl observed.
• (d) N-methylpropanamide formed; excess methylamine is needed so the second mole can react with the HCl by-product, forming methylammonium chloride.