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Is this reaction $\ce{S_N2}$ (because attack of nucleophile and removal of leaving group are not taking place in a single step)?

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Which part of anhydride should form the substituted amide? If it is an $\ce{S_N2}$, then $\ce{CH3COO-}$ is a better leaving group but a more (alkyl) $\ce{e-}$ releasing group, $\ce{-C2H5}$ would destabilize the T.S.

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The reaction between anhydride and primary amine is by the Nucleophic addition-elimination mechanism (NOT by SN2).

First step: Nucleophilic addition step (Rate determining step):

The addition of amine to carbonyl center (at the most electrophilic carbonyl center and it should generate least sterically hindered intermediate) gives rise to a tetrahedral intermediate (4 groups bonded to the carbon center), which is influenced by steric hinderance. Therefore the attack of primary amine will prominently take place at -CH3 end carbonyl center which is more electrophilic and gives less sterically hindered intermediate.

Second step: Elimination step (fast step, not a rate determining):

The internal nucleophile (O- of earlier carbonyl) reverts back to generate the carbonyl center and propionate ion is eliminated.

Note: As the mechanism of reaction is not by SN2, therefore leaving group stability is not the sole criteria.

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    $\begingroup$ There's no such thing as a "sterically hindered intermediate". Steric hindrance is a kinetic property of a reaction, not a thermodynamic property of a molecule. $\endgroup$
    – orthocresol
    Nov 11 '21 at 16:52
  • $\begingroup$ Then "sterically hindered transition state or pathway" will be correct ? $\endgroup$
    – Sandee
    Nov 11 '21 at 19:12
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    $\begingroup$ Pathway seems like a better choice. $\endgroup$
    – orthocresol
    Nov 11 '21 at 19:55

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