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34.3 Amides

Syllabus
9701–2028–2029
Topic
34.3
Level
A2

Learning objectives

34.3.1Reactions (reagents/conditions)• Recall the reactions (reagents/conditions) by which amides are produced:: (a) the reaction between ammonia and an acyl chloride at room temp.; (b) the reaction between a primary amine and an acyl chloride at room temp.34.3.2Reactions of amides• Describe reactions of amides:: (a) hydrolysis with aq. alkali or aq. acid; (b) the reduction of the CO group in amides with LiAl H₄ to form an amine34.3.3Why amides are much weaker bases than amines• State/explain: why amides are much weaker bases than amines

Ammonia and primary amines form amides with acyl chlorides at room temperature

At room temperature, ammonia reacts with an acyl chloride to form an unsubstituted amide, while a primary amine forms an N-substituted amide. In each condensation, the nitrogen-containing nucleophile replaces Cl at the acyl carbon and HCl is eliminated.

Nitrogen reactant Acyl chloride Amide product Acid captured by excess reactant
ammonia, NH3 RCOCl RCONH2 NH4Cl
primary amine, R'NH2 RCOCl RCONHR' R'NH3Cl

RCOCl+2 NHX3→RCONHX2+NHX4Cl\ce{RCOCl + 2NH3 -> RCONH2 + NH4Cl}

RCOCl+2 RX′NHX2→RCONHRX′+RX′NHX3Cl\ce{RCOCl + 2R'NH2 -> RCONHR' + R'NH3Cl}

Only one ammonia or amine molecule supplies the nitrogen in the amide; the second equivalent captures HCl. Do not include secondary amines in this specific 34.3.1 recall objective, which names ammonia and primary amines.

Amides undergo hydrolysis or C=O reduction with different structural outcomes

Hydrolysis breaks the acyl C-N bond and separates the carbonyl and nitrogen fragments. LiAlH4 reduction keeps the C-N skeleton together but changes the amide carbonyl carbon into CH2, forming an amine.

Amide reaction Reagents and conditions Carbonyl-side product Nitrogen-side product
acid hydrolysis aqueous acid, heat/reflux carboxylic acid, RCOOH NH4+ from RCONH2, or R'NH3+ from RCONHR'
alkaline hydrolysis aqueous alkali, heat/reflux carboxylate, RCOO- NH3 from RCONH2, or R'NH2 from RCONHR'
reduction LiAlH4 in dry ether, then work-up C=O becomes CH2; no separate carbonyl fragment RCONH2 -> RCH2NH2; RCONHR' -> RCH2NHR'

RCONHX2+HX2O+HX+→RCOOH+NHX4X+\ce{RCONH2 + H2O + H+ -> RCOOH + NH4+}

RCONHX2+OHX−→RCOOX−+NHX3\ce{RCONH2 + OH- -> RCOO- + NH3}

In LiAlH4 reduction, retain the former carbonyl carbon: ethanamide, CH3CONH2, forms ethylamine, CH3CH2NH2. Reduction does not remove that carbon or split the C-N bond.

Do not write identical hydrolysis products for acid and alkali. Acid protonates ammonia/amine; alkali deprotonates the carboxylic acid to a carboxylate. Reduction is a separate reaction and gives an amine without hydrolytic cleavage.

Amides are much weaker bases because the nitrogen lone pair is delocalised

An amine nitrogen lone pair is relatively localised and available to accept H+. In an amide, the nitrogen lone pair overlaps with the adjacent C=O pi system and is delocalised across the O-C-N unit.

This delocalisation gives the C-N bond partial double-bond character and stabilises the unprotonated amide. Using the nitrogen lone pair to bond to H+ would remove it from that conjugated system, so it is much less available than an amine lone pair.

basicity: amine≫amide\text{basicity: amine} \gg \text{amide}

The amide nitrogen still has a lone pair, but presence is not the same as availability. Do not explain the difference only by saying that both compounds contain nitrogen; compare what happens to the lone pair next to C=O.