Methylamine, CH3NH2, is reacted with propanoyl chloride, CH3CH2COCl, at room temperature. (a) Give an equation for this reaction, and name the amide formed. [2] (b) State and explain why this amide is a much weaker base than methylamine. [3] (c) The amide is then heated under re…

Answer: ((a)) CH3CH2COCl + CH3NH2 → CH3CH2CONHCH3 + HCl; the amide formed is N-methylpropanamide. ((b)) The nitrogen lone pair in the amide overlaps with (delocalises into) the adjacent C=O pi system, so it is far less available to accept a proton than the localised lone pair on methylamine's nitrogen; this makes the amide a much weaker base than methylamine. ((c)) Propanoic acid, CH3CH2COOH, and methylammonium chloride, CH3NH3^+Cl^-.. Part (a): Forming the amide A primary amine reacts with an acyl chloride at room temperature in a condensation reaction, forming an amide and releasing HCl: CH3CH2COCl + CH3NH2 → CH3CH2CONHCH3 + HCl The nitrogen atom of methylamine's lone…

A Level Chemistry 9701: Amide Formation, Basicity and Hydrolysis

Part (a): Forming the amide

A primary amine reacts with an acyl chloride at room temperature in a condensation reaction, forming an amide and releasing HCl:

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

The nitrogen atom of methylamine’s lone pair attacks the carbonyl carbon of propanoyl chloride, displacing chloride and forming a new C–N bond. The organic product is N-methylpropanamide: “propan-” for the three-carbon chain from propanoyl chloride, “-amide” for the $-\text{CONH}-$ linkage, and “N-methyl” to show that the methyl group is attached to the amide nitrogen rather than to the carbon chain.

Part (b): Why the amide is a much weaker base than methylamine

Methylamine is a reasonably strong base because its nitrogen lone pair is localised on the nitrogen atom and is free to accept a proton.

In N-methylpropanamide, however, the nitrogen atom is directly bonded to the carbonyl carbon of the $\text{C=O}$ group. The nitrogen lone pair overlaps with (delocalises into) this carbonyl pi system, similarly to how phenylamine’s lone pair delocalises into a benzene ring. This delocalisation:

spreads the lone pair’s electron density away from the nitrogen atom, so it is much less available to bond to an incoming $\text{H}^+$ ion, and
would be lost if the lone pair were used to accept a proton, making protonation energetically unfavourable.

For these reasons, the amide is a much weaker base than the amine it was made from, so much weaker that aqueous amide solutions are essentially neutral.

Part (c): Hydrolysing the amide with dilute acid

Heating the amide under reflux with dilute hydrochloric acid hydrolyses the amide bond. Water adds across the $-\text{CONH}-$ linkage, and because the solution is acidic, the amine product is protonated as it is released:

$\text{CH}_3\text{CH}_2\text{CONHCH}_3 + \text{H}_2\text{O} + \text{HCl} \rightarrow \text{CH}_3\text{CH}_2\text{COOH} + \text{CH}_3\text{NH}_3^+\text{Cl}^-$

The two products are propanoic acid, $\text{CH}_3\text{CH}_2\text{COOH}$ , and methylammonium chloride, $\text{CH}_3\text{NH}_3^+\text{Cl}^-$ (methylamine exists as its protonated ammonium salt under these acidic conditions, rather than as free methylamine).

Final answers

(a) $\text{CH}_3\text{CH}_2\text{COCl} + \text{CH}_3\text{NH}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CONHCH}_3 + \text{HCl}$ ; N-methylpropanamide.
(b) Much weaker base: the amide nitrogen lone pair is delocalised into the adjacent C=O group, making it far less available to accept a proton than the localised lone pair on methylamine.
(c) Propanoic acid, $\text{CH}_3\text{CH}_2\text{COOH}$ , and methylammonium chloride, $\text{CH}_3\text{NH}_3^+\text{Cl}^-$ .