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Amides require much harsher conditions to hydrolyse than its ester homologue. An explanation given is that the orbitals holding the lone pair on the nitrogen overlaps with the C=O π-bond to give conjugation, thus introducing a "partial" π-bond between nitrogen and the carbonyl carbon. Consequently, this makes the bond harder to break.

However, for the ester homologue, doesn't this conjugation also happen? An explanation given is poorer overlapping of the orbitals. However, how is that true?

In addition, such explanation is given to show that C=O is shorter in ester than it is in amide. However, I still don't get why this is the case.

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  • $\begingroup$ Amide = crummy leaving group, maybe? $\endgroup$ Nov 30 '20 at 18:05
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    $\begingroup$ If harsh conditions are needed because of kinetics rather than thermodynamics, the main question is: what is the rate-determining step (rds)? Anything that facilitates that step should make the reaction faster. For acid-catalysed amide hydrolysis, it appears that the rds is the attack of water. Compare a protonated ester and a protonated amide: which one has the highest electron-density at the carbon that water must attack? See also this. $\endgroup$ Nov 30 '20 at 19:15
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Dimethyl acetamide (DMA) is less susceptible to hydrolysis than ethyl acetate (EA). This statement is true for aqueous acid or base. The infrared (IR) frequency for the carbonyl group of DMA is 1662 cm-1 while EA is 1742 cm-1. The lower the frequency, the lower the energy and the more, relatively speaking, single bond character in the carbonyl group. Thus, the carbonyl group of DMA has less double bond character than that of EA. Because nitrogen can sustain a positive charge better than oxygen, resonance is more pronounced in DMA than in EA. Because resonance is stronger in DMA, its carbonyl group is less susceptible to nucleophilic attack. The carbonyl bond length of EA would be expected to be shorter than that for DMA because the more double bond character, the shorter the bond. It is no different than the carbon-carbon bond length in the series ethane, ethylene and acetylene.

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    $\begingroup$ Please would you elaborate on why nitrogen can better sustain the charge than oxygen, if oxygen has higher electronegativity? In this case, nitrogen is better because of the inductive effect due to the 2 methyl groups, and also possible sigma conjugation? would you explain by comparing a primary amide with its homologue? Thanks @user55119 $\endgroup$
    – Yushi Li
    Nov 30 '20 at 20:31
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    $\begingroup$ The methyl groups do not have to be present. Amides are less reactive than esters in general. The more electronegative, oxygen, is less willing to bear a positive charge than nitrogen. CH3OCH2Cl is a liquid that reacts with nucleophiles by prior ionization to [CH3O=CH2]+ Cl-. OTOH, Me2NCH2Cl exists as a salt, [Me2N=CH2]+ Cl-. Now you can explain why acyl chlorides are more reactive than esters and amides. $\endgroup$
    – user55119
    Nov 30 '20 at 20:45

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